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Space Complexity - (Formal Verification of Hardware) - Vocab, Definition, Explanations | Fiveable
Space Complexity
from class:
Formal Verification of Hardware
Space complexity refers to the amount of memory required by an algorithm to run as a function of the length of the input. This includes both the temporary space allocated by the algorithm during its
execution and the space required for the input itself. Understanding space complexity is crucial because it helps evaluate the efficiency of algorithms, particularly when working with large data sets
or in resource-constrained environments.
congrats on reading the definition of Space Complexity. now let's actually learn it.
5 Must Know Facts For Your Next Test
1. Space complexity is typically expressed using Big O notation, which helps categorize algorithms based on their memory usage relative to input size.
2. The total space complexity can be broken down into two parts: fixed part (space required for constants, simple variables, fixed-size variables) and variable part (space required by dynamically
allocated variables and function call stack).
3. Understanding space complexity is essential when optimizing algorithms for performance, especially in systems with limited memory resources.
4. Recursive algorithms often have higher space complexity due to additional stack space required for each function call.
5. In certain applications, such as embedded systems or mobile devices, minimizing space complexity can be as critical as minimizing time complexity.
Review Questions
• How does space complexity influence the choice of algorithms in practical applications?
□ Space complexity influences algorithm selection by determining how much memory will be consumed as inputs increase. In applications where memory resources are constrained, such as mobile
devices or embedded systems, algorithms with lower space complexity are preferred. Additionally, understanding space complexity allows developers to avoid potential memory overflow issues and
optimize resource usage effectively.
• Compare and contrast time complexity and space complexity in terms of their importance for algorithm design.
□ Time complexity and space complexity are both critical metrics for evaluating algorithms, but they focus on different resources. While time complexity looks at how long an algorithm takes to
execute based on input size, space complexity assesses how much memory it uses. In many cases, there is a trade-off; optimizing for speed may increase memory use and vice versa. A good
algorithm design balances both complexities according to the application's needs.
• Evaluate how recursive algorithms may impact overall space complexity compared to iterative solutions.
□ Recursive algorithms typically have a higher overall space complexity than iterative solutions due to the additional stack space consumed with each function call. Each recursive call requires
memory for parameters and local variables, which adds up quickly with deep recursion. In contrast, iterative solutions reuse the same variables within a loop structure, leading to lower
memory usage. Understanding these differences helps developers choose appropriate strategies based on performance requirements and system limitations.
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Endpoint Math Formula | Synonym
Hemera Technologies/AbleStock.com/Getty Images
The line is one of the simplest mathematical objects. Even so, finding all the information about a single line sometimes requires some number-crunching. When you know partial information about your
line, you will often have to plug your known information into formulas to describe the rest of the line. The endpoint formula is one such formula, and it can help you find the unknown endpoint of an
otherwise well-understood line.
1 The Problem: Where’s the Rest of My Line?
Often in math, you are left with pieces of information with which you must piece together the entire situation. This often occurs in geometry, a subject in which you must infer conclusions about
lines and other shapes based on partial information about them. In some cases, you might be looking at information describing a line segment, but are missing the endpoint of the line. A line segment
is defined by its two endpoints, the knowledge of which allows you to fully describe the line segment in mathematics. Thus, you need to know both endpoints before you can clearly label or define a
line segment.
2 The Midpoint Formula: Half of the Story
Knowing the midpoint formula for a line segment will often help you derive the endpoint. Recall the midpoint formula being the mathematical formula that allows you to find the point lying in the
center of a line segment, based on the two endpoints. Specifically, if your endpoints are (x0, y0) and (x2, y2), the midpoint will have an x-coordinate at (x0+x2)/2 and a y-coordinate at (y0+y2)/2.
By plugging in the endpoints into the midpoint formula, you can find the midpoint of the segment. But what many students don’t notice is that you can reverse the midpoint equation to find the
3 The Endpoint Formula: The Rest of the Story
If you have one endpoint and the segment’s midpoint, you can apply a modification of the midpoint formula to derive the other endpoint. The midpoint formula lets you find a coordinate by plugging the
information you know into p1 = (p0 + p2)/2. If you don’t know one of the endpoints, either p0 or p2, but do know the midpoint, p1, you can solve this equation for the other endpoint. Using algebra,
you can rewrite the midpoint formula as 2_p1-p0. Thus, your endpoint has an x-coordinate at 2_x1-x0 and a y-coordinate at 2*y1-y0, where (x1,y1) is the midpoint of the line segment and (x0,y0) is the
endpoint of the segment.
4 A Quick Example
Assume you have a line segment with an endpoint at (-2,7) and a midpoint at (-9,2). You want to find the other endpoint through the endpoint formula, so you first label your variables: x0=-2, y0=7,
x1=-9, and y1=2. Find the x-coordinate first: x2=2_x1-x0=2_(-9)-(-2)=-16. Find the y-coordinate next: y2=2_y1-y0=2_2-7=-3. Thus, your other endpoint is at (-16,-3). | {"url":"https://classroom.synonym.com/endpoint-math-formula-33008.html","timestamp":"2024-11-04T23:29:35Z","content_type":"text/html","content_length":"240690","record_id":"<urn:uuid:a733afcb-9122-4fcd-af9b-7bba4417f5dc>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.84/warc/CC-MAIN-20241104225856-20241105015856-00457.warc.gz"} |
Rules for significant figures in Calculations | Edumir-Physics
Rules for significant figures in Calculations
Suppose you have calculated a value and the calculator is showing it as 10.333333…. Now, what will you write to express the number? Up to what decimal number will you take the data? To know this, the
concept of significant digits is essential. Significant figures or significant digits are very important in measurements to write the data in a systematic way. In this article, we’re going to learn
some rules of significant figures in calculations and a systematic way to present a number up to any significant figure.
Contents in this article:
1. What are the significant figures?
2. Importance of significant figures
3. Rules for significant figures
4. Rules for significant figures in calculations
5. Calculation of Significant figures for addition
6. Significant figures for subtraction
7. Significant figures in Multiplication
8. Division in significant figures
What are the significant figures?
Significant figures are the digits in a number that are used to increase the precision of the calculation in a measurement. The Significant figure is also known as significant digit. It has an
important role to make the calculated data precise with the measured data.
Importance of significant figures in Calculations
Significant figures in calculations have an important role to match the calculated values with the measured values with greater precision. Therefore, we need to know the rules to find the number of
significant figures in a given number. One can understand this by the following example –
Let in a circuit a 3 ohm resistance is connected with a 10 Volt battery. The current through the circuit can be measured with an ammeter. Let the smallest division of the ammeter is 0.01 ampere and
it measures the current as 3.33 ampere. So, the measured value is 3.33 ampere.
Again, one can calculate the value of current flow through the resistance by Ohm’s law of current electricity. Current, I=V/R which gives the calculated value of the current as 3.333333333… ampere.
Clearly, the calculated value is not precise with the measured value. If we take the calculated value in two significant digits after the decimal then the calculated value also be 3.33 Ampere. Then
it becomes precise with the measured value. Therefore, significant digits are important in measurements and calculations.
Rules with significant figures
There are some rules to find the number of significant figures in a given number. Here, we are going to discuss those rules with examples. We will write the significant digits in green color and
non-significant digits in red color. To understand the whole things we take three random numbers – i) 0.00250 ii) 2.00530 and iii) 6.022×10^23
1. All the Non-zero digits are significant. So, in 0.00250 the digits 2 and 5 are significant figures. In 2.00530 the digits 2, 5 and 3 are significant figures. But there may be more significant
digits among the zeros. we will learn those in the next rules.
2. All the zeros at the beginning (left side) of a number are not significant. So, in the number 0.00250, the first three zeros are not significant. But in other two numbers, there are no zeros at
the beginning.
3. The zero between two significant digits is significant. So, in the number 2.00530, the green zeros between 2 and 5 are significant digits. Also in 6.022, the zero is between two significant
digits 6 and 2. So, zero in 6.022 is also significant.
4. Any zero at the end (at the extreme right) after the decimal is significant. So, in 0.00250 and in 2.00530 the green zeros are significant. But the zero at the extreme right of a whole number is
not significant. Example: 1 and 2 are the only two significant figures in 1200.
5. Power of 10 is not significant. In the Avogadro number 6.022×10^23, the power of 10 is not a significant figure.
So the three numbers can be expressed in significant form as i) 0.00250 ii) 2.00530 and iii) 6.022×10^23
Clearly, the number 0.00250 has three significant digits, 2.00530 has six significant digits and there are four significant digits in 6.022×10^23.
Homework problems: Find the number of significant figures in the numbers i) 0.250800 ii) 100 iii) 110.0070
Rules for significant figures in calculations
From the first paragraph, you learn the importance of significant figures in calculations. But how to use significant digits in calculations like addition, subtraction, Multiplication and division.
These are discussed below. Before this, you should have the concept of rounding off the numbers. Read this post on Rounding off significant figures.
Significant figures for addition
For the addition or subtraction of two or more numbers, one should concentrate on the significant digits after the decimal. In both cases, add or subtract the numbers in the usual manner. Then
rounding off the result up to the lowest number of significant figures that any of those numbers has.
Example: Add the numbers 2.30, 2.578 and 2.5 to the significant figure.
General addition of those numbers is = (2.30+2.578+2.5)= 7.378
Now among the three numbers 2.5 has one, 2.30 has two and 2.578 has three significant figures respectively after the decimal. So the result of the addition should be up to one significant digit
(lowest one) after the decimal. So, the answer to the addition in significant figure will be 7.4 (after rounding off).
Rules for Significant figures in Subtraction
Rules: First, do the usual subtraction. Then round off the result in the smallest significant figures that any of the given numbers have. Focus on significant digits after the decimal only.
Example: Subtract 1.72 from 5.218 in significant figures.
Subtraction of 1.72 from 5.218 is = (5.218 – 1.72) = 3.498
Now 1.72 has the smallest number of significant digits after the decimal. So, the answer will be 3.50 in significant figures after the rounding off.
Significant figures for multiplication
Rules: Do the usual multiplication and then write the result in the form of the smallest number of significant figures that the given numbers have.
Example-1: Multiply 3 by 2.5 in the significant figure.
The multiplication between 3 and 2.5 is 7.5.
Now the number 3 has the smallest significant figure after the decimal which is nothing. So the result will not contain any digit after the decimal. Therefore, the answer is 8 as 7 is odd any after 7
there is 5.
Emaple-2: Multiply 5 by 2.5 in the significant figure.
The multiplication of 5 with 2.5 is 12.5. But among the given numbers 5 has no digit after the decimal. So, the answer will be 12.
Rules for significant figures when dividing
Rules: Perform the usual division and then write the result in the form the smallest number of significant figures that the given numbers have.
Example: Divide 5.5 by 3.11 in a significant figure.
The division of 5.5 by 3.11 is 1.7684… Now, among the given numbers, 5.5 has one digit after the decimal and 3.11 has two digits after the decimal. So, the answer should contain the minimum digit
after the decimal which is up to one. Hence the answer is 1.8 after rounding off.
Solved problems on significant figures
1. Find the number of significant figures in the number 1.065. Answer: In this case, all the digits are significant. Check rule-1 and 3 in above. So, there are 4 significant figures in this number.
2. Find the number of significant figures in 0.06900. Answer: Here, the digits 6, 9 and the two zeros at the extreme right are significant figures. Follow rule-1, 2, 4 and 3 above to figure it out.
So, the number of significant figures in 0.06900 is four.
3. Find the number of significant figures in 100. Answer: 100 is a whole number. Rule-4 above says that zeros at the extreme right in a whole number are not significant. So, the number 100 has only
one significant figure which is 1.
Homework problems:
1. Add 2.76 and 4.995 in significant figures
2. Multiply 10.2 with 0.9 in significant figures.
In this article, we learned the rules for significant figures in calculations. This is all from this article. If you have any doubts on this topic you may ask me in the comment section.
Thank you!
Related posts:
To check all physics-related posts, visit the homepage.
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How to highlight the values in the column which is greater than or less than the other column values
Suppose, the order quantity we have taken from the customer is 500, and the quantity we have placed an order with our supplier is 600 how to highlight this excess quantity?
Best Answer
• Hello @RUPESH KUMAR
One option is to set up a helper column for symbols (such as traffic lights).
In this column you'll enter a formula something like:
=IF(Ordered@row > [Cust. Order Amount]@row, "Yes", IF(Ordered@row < [Cust. Order Amount]@row, "No", "Hold"))
You'll then use the Conditional Formatting button in the tool bar (blue box in picture above) and set up similar to this:
I have only chosen to highlight the Customer Order Amount cell. You can change it to the Ordered cell, other cells, or the whole row.
Hope this helps and that you have a great day,
Jason Albrecht MBA, MBus(AppFin), DipFinMgt
LinkedIn profile - Open to work
• Hello @RUPESH KUMAR
One option is to set up a helper column for symbols (such as traffic lights).
In this column you'll enter a formula something like:
=IF(Ordered@row > [Cust. Order Amount]@row, "Yes", IF(Ordered@row < [Cust. Order Amount]@row, "No", "Hold"))
You'll then use the Conditional Formatting button in the tool bar (blue box in picture above) and set up similar to this:
I have only chosen to highlight the Customer Order Amount cell. You can change it to the Ordered cell, other cells, or the whole row.
Hope this helps and that you have a great day,
Jason Albrecht MBA, MBus(AppFin), DipFinMgt
LinkedIn profile - Open to work
Help Article Resources | {"url":"https://community.smartsheet.com/discussion/118506/how-to-highlight-the-values-in-the-column-which-is-greater-than-or-less-than-the-other-column-values","timestamp":"2024-11-06T21:29:50Z","content_type":"text/html","content_length":"407814","record_id":"<urn:uuid:2d55c2ad-0396-4262-be44-f66ce9b1ce97>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00058.warc.gz"} |
manual pages
Hierarchical random graphs
Fitting and sampling hierarchical random graph models.
A hierarchical random graph is an ensemble of undirected graphs with n vertices. It is defined via a binary tree with n leaf and n-1 internal vertices, where the internal vertices are labeled with
probabilities. The probability that two vertices are connected in the random graph is given by the probability label at their closest common ancestor.
Please see references below for more about hierarchical random graphs.
igraph contains functions for fitting HRG models to a given network (fit_hrg, for generating networks from a given HRG ensemble (sample_hrg), converting an igraph graph to a HRG and back (hrg,
hrg_tree), for calculating a consensus tree from a set of sampled HRGs (consensus_tree) and for predicting missing edges in a network based on its HRG models (predict_edges).
The igraph HRG implementation is heavily based on the code published by Aaron Clauset, at his website (not functional any more).
See Also
Other hierarchical random graph functions: consensus_tree(), fit_hrg(), hrg_tree(), hrg(), predict_edges(), print.igraphHRGConsensus(), print.igraphHRG(), sample_hrg()
version 1.2.5 | {"url":"https://igraph.org/r/html/1.2.5/hrg-methods.html","timestamp":"2024-11-01T22:33:51Z","content_type":"text/html","content_length":"9867","record_id":"<urn:uuid:214e7107-f8bd-49dd-b10f-9d8c77a23deb>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00473.warc.gz"} |
Rigorous scaling laws for internally heated convection at infinite Prandtl number
Arslan A, Fantuzzi G, John C, Wynn A (2023)
Publication Language: English
Publication Status: In review
Publication Type: Journal article, Original article
Future Publication Type: Journal article
Publication year: 2023
Book Volume: 64
Article Number: 023101
DOI: 10.1063/5.0098250
Open Access Link: https://arxiv.org/abs/2205.03175
New bounds are proven on the mean vertical convective heat transport, \overline{⟨wT⟩}, for uniform internally heated (IH) convection in the limit of infinite Prandtl number. For fluid in a
horizontally-periodic layer between isothermal boundaries, we show that \overline{⟨wT⟩}≤1/2−cR^−2, where R is a nondimensional `flux' Rayleigh number quantifying the strength of internal heating and
c=216. Then, \overline{⟨wT⟩}=0 corresponds to vertical heat transport by conduction alone, while \overline{⟨wT⟩}>0 represents the enhancement of vertical heat transport upwards due to convective
motion. If, instead, the lower boundary is a thermal insulator, then we obtain \overline{⟨wT⟩}≤1/2−cR^−4, with c≈0.0107. This result implies that the Nusselt number Nu, defined as the ratio of the
total-to-conductive heat transport, satisfies Nu≲R4. Both bounds are obtained by combining the background method with a minimum principle for the fluid's temperature and with Hardy-Rellich
inequalities to exploit the link between the vertical velocity and temperature. In both cases, power-law dependence on R improves the previously best-known bounds, which, although valid at both
infinite and finite Prandtl numbers, approach the uniform bound exponentially with R.
Authors with CRIS profile
Additional Organisation(s)
How to cite
Arslan, A., Fantuzzi, G., John, C., & Wynn, A. (2023). Rigorous scaling laws for internally heated convection at infinite Prandtl number. Journal of Mathematical Physics, 64. https://doi.org/10.1063/
Arslan, Ali, et al. "Rigorous scaling laws for internally heated convection at infinite Prandtl number." Journal of Mathematical Physics 64 (2023).
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MATHEMATICS- GEOMETRICAL EQUATIONS - Great Rock Dev
Since old times, mathematics has been an indispensable part of various societies such as physical sciences and technological works. In recent times, mathematics has gained a good amount of
significance in other fields too. This significance can be elucidated by the fact that mathematics is a study of geometry, algebra, numbers, and various shapes. One of the specific aspects of this
subject includes geometry.
Geometry is the learning of structure and knowledge of different shapes and equations. Knowledge of these shapes includes the area, perimeter, and several other key points about these figures. The
concept of geometry also includes the topic of circles, lines, and equations. A circle as it is known is a round plane figure whose every point on the circumference is equidistant to the center fixed
point. On the other hand, the line is a straight length that can be extended from either side and consists of no width. Both of these structures have abundant topics to understand. Here, we are going
to learn and expertise the basics of the equation of line and equation of a circle. Now let us discuss this concept more.
Equations of a circle
The equation of a circle can be defined as an algebraic expression of the circle located in the XY plane. This equation helps us in recognizing the center of the circle and the measure of the radius.
The Basic equation of circle is recorded as:
(x-h)2 + (y-k)2 = r2,
where h and k are the fixed center points and r is referred to as the center.
Equations of a line
Equation of line can be elaborated as a geometric equation that expresses the length and structure of line in an algebraic expression that is easily understandable to be located in an XY plane. The
standard equation of a line in two variables is:
Ax + By +C = 0.
Where, A, B, and C are constants and A, B are not equal to zero. These equations can also be converted to a linear equation in a single or three variables.
There are some other equations of circles as well except the standard equation. These algebraic expressions exhibit different situations like the equation of a circle when the center is the origin,
the equation when the center is not the origin. In the same way, there is some other equation of line as well. To illustrate- the standard forms of the linear equation can be differentiated into
three parts. These are slope-intercept form, intercept form, and normal form. These equations are further known as straight-line formulas.
Coming to an end, this was a basic commencement about the concepts of equations of lines and circles whereas, this does include plenty of content to study, understand and learn. These are vast
concepts and require needful practice and effort. It results in several queries and doubts in every student.
Cuemath is an online tutoring center that provides you with highly qualified and experienced teachers. They provide one-to-one attention to every student in personal sessions to clear all the
queries. This assists in the efficient, effective, and mindful growth of students. Another advantage of these classes is that they are online. Anyone from anywhere can join these classes.
It benefits everyone as you won’t have to travel and can learn everything from the comfort of your home. Moreover, they also provide coding lectures which help in the enhancement and growth of
students. Hence, joining these classes can result in the positive development of the student and benefit them well. | {"url":"https://greatrockdev.com/mathematics-geometrical-equations/","timestamp":"2024-11-04T06:06:30Z","content_type":"text/html","content_length":"67479","record_id":"<urn:uuid:a3c783c6-942d-425a-a332-007d074e9036>","cc-path":"CC-MAIN-2024-46/segments/1730477027812.67/warc/CC-MAIN-20241104034319-20241104064319-00213.warc.gz"} |
medical tests and probabilities
a mini post
There are many references in the news to how good viruses tests are and what conclusions you can draw from them. Unfortunately some are wrong and I haven't seen any easy to understand explanations
of how to figure these things out. It's all conditional probability. The standard approach is to use Bayes' Theorem. Unfortunately most people have never heard of it or vaguely recall a formula
with P's and |'s they once memorized without knowing what it really did
Here's an easy way to think about it in the form of an example. Imagine a population, say a city, where 1% of the people have a viral disease (I'll call the disease 'virus'). There's a test with a
90% chance of returning a correct result. So if you have the disease there's a 90% chance the test returns a positive, and if you don't have the test there's a 90% chance the test returns a
negative. So what's the probability that someone who has just been tested has the disease?
For simplicity imagine an unbiased group of 1000 people in the city. Let's screen them and see what happens. Ten people will have the disease, 990 will be disease-free.Of the ten people with the
disease the test will correctly identify nine of them and incorrectly identify one. Looking at the people who do not have the disease 90% of 990 or 891 will correctly test negative, but 10% of the
990 or 99 will incorrectly test positive.
So the probability a person who tests positive actually has the disease is the number who test positive and have the disease divided by the total number who test positive: 9/108 or about about an 8%
The example has nothing to do with the specifics of CV-19, but maybe will give you a way to think about how to sort out what the real tests are saying when an accuracy is published. Going deeper
really requires tools like Bayes' theorem, but hopefully this is clear.
Armed with this technique perhaps this otherwise well-written piece in the NY Times will make sense. Also note this is for an unbiased initial sample. If you're screening on existing conditions
before testing (which is common outside of Iceland), additional complications appear. But at least this should give you an idea.
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Please help me | Sololearn: Learn to code for FREE!
Please help me
Howto solve this problem in python ??? 2. This is a program to analyze mobile numbers using match() function. Note: re.compile() complies a regular expression pattern into a regular expression
object, which can be used for matching using its match() or search(). Mobile phone number in Korea starts with 3 digit numbers followed by 3~4 and 4 digits fixed customer numbers. (e.g.,
010-2345-5678 or 010-234-5678) <Output> ('010', '123', '45
use this x=re.compile(r'(\d{3} | \d{4})') | {"url":"https://www.sololearn.com/en/discuss/1212155/please-help-me","timestamp":"2024-11-06T00:55:12Z","content_type":"text/html","content_length":"913289","record_id":"<urn:uuid:5e238ea9-e15b-4e1d-bc13-e2ce0f3ab833>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00584.warc.gz"} |
Strategic Placement of Charging Stations for Enhanced Electric Vehicle Adoption in San Diego, California
Strategic Placement of Charging Stations for Enhanced Electric Vehicle Adoption in San Diego, California ()
1. Introduction
California Governor Gavin Newsom has mandated that 100% of cars bought and sold in California must be electric by 2035 [1] . Other states will likely follow California’s bold lead toward electrifying
the car market. The current standard model for electric vehicle (EV) charging relies on the driver having access to a permanent parking spot or garage for nightly charging. As EVs reach higher
penetration, urban residents will need a way to charge their vehicles, despite reliance on street parking or an inconsistent rotation of available spots. To meet the charging needs of its residents,
cities must think critically about placing publicly accessible charging stations such that there is equitable access to all residents. We define equality as guaranteeing a base local supply for all
areas; in contrast, we define equity as maintaining consistent local supply across all areas, including excess supply. Equity and equality in this report are in terms of distance to a charging
station and number of available chargers. We will be investigating San Diego County because it has well-defined geographical boundaries [2] and its residents rely heavily on personal vehicles for
transportation [3] .
Our objective function represents a sum of all charging supplies. Therefore, we are minimizing the capacity of charging needed to meet demand, such that all people have close access to a charger. We
assume that charging demand is concentrated at the centroids of the census block groups in the county, which we call population centers. The willingness to drive a certain distance from the center is
modeled as a Gaussian function to determine how well a charging station can serve that population center. This population center approach leads to a non-convex objective function, as there are
approximately 1796 census block groups in San Diego County. A trivial result of this optimization is that there is a single charging station at every population center, with a number of chargers that
are proportional to the demand at that point. This is not an optimal solution, as it would require a minimum of 1796 chargers. The charger supply would be greater than the demand and purchasing that
many chargers is also economically infeasible. Hence, we will assign and parameterize a number of charging stations that is lower than the number of population centers.
To test the formulation of our algorithm and parameterize the number of charging stations we have used Coronado Island as a ’toy problem’ or smaller test case for our algorithm. This allows for
easier comparison of charging station placement. To meet demand most efficiently, we add the constraints that the demand at each population center is met and that the demand on each charging station
does not exceed supply. Once we have optimized to meet demand, we will apply a third constraint that the station is on a road. This will require a second round of optimization to determine the road
vertex that is closest to the charging station location. We are using publicly available Geographical Information Systems (GIS) data to get location-based information for variables and parameters
such as population density and road locations [3] [4] .
2. Problem Statement and Analysis
For this investigation, we want to determine the minimum amount of geographically weighted supply (i.e. charging spots) that can still satisfy all geographically weighted demand. We did so by
modeling the problem as a flow network shown in Figure 1 where the supply from charging stations feeds into the demand from population centers, with the weights determined by distance. Table 1 shows
Table 1. Model variables and parameters.
different types of model variable and parameters used for the calculation along with their units.
$f\left(x\right)=\frac{{\sum }_{i=1}^{N}\text{\hspace{0.17em}}{x}_{n,i}}{{\sum }_{j=1}^{{N}_{P}}\text{\hspace{0.17em}}{P}_{j}}$ Station supply scaled by demand
With Respect to:
$x={\left[{x}_{lat},{x}_{long},{x}_{n}\right]}^{\text{T}}\in {ℝ}^{N}$ Latitude, longitude & station supply
Subject to:
${h}_{1,i}\left(x\right)={\left[\begin{array}{c}{x}_{lat}\\ {x}_{long}\end{array}\right]}_{i}-{R}_{i}{}_{2}=0$${\forall }_{i}$ Station is on road.
${g}_{1,j}\left(x\right)=\frac{1}{10}\left({o}_{j}-{\sum }_{i=1}^{N}{x}_{n,i}{w}_{ij}\right)\le 0$${\forall }_{j}$ Demand at each population center is met.
${g}_{2,i}\left(x\right)={\sum }_{j=1}^{{N}_{P}}{w}_{ij}-1\le 0$${\forall }_{j}$ Station supply must at least meet demand.
${w}_{ij}=\mathrm{exp}\left(-{\left[\begin{array}{c}{x}_{long}\\ {x}_{lat}\end{array}\right]}_{i}-{P}_{j}{}_{2}^{2}\frac{1}{\beta }\right)$${\forall }_{i,j}$ Normalized willingness to drive to
$32.58\le {x}_{lat}\le 32.72$ Geographic bounds of Coronado Island
$-117.23\le {x}_{long}\le -117.12$
$i\in \left\{1,2,\cdots ,N\right\},j\in \left\{1,2,\cdots ,{N}_{P}\right\}$
2.1. Assumptions
Several assumptions were made to simplify the problem and arrive at a feasible solution. All the assumptions are valid in the feasible domain. These are listed below.
· Optimized station locations are snapped to the closest road assuming that it still sufficiently satisfies the demand.
· Existing EV charging infrastructure has not been taken into consideration.
· All vehicles are assumed to be EVs compatible with Level 3 charging.
· Vehicle charging demand is assumed to be linearly dependent on the population and the county-wide car/population ratio of 0.73 approximately.
· A household is assumed to be the central unit for the demand analysis where each household consists of 2.73 people and two cars on average [5] . See Appendix A for further details.
· Willingness to travel to a public charger is modeled as a Gaussian distribution centered around the population center and having a variance of 0.00005 (β = 0.0001).
2.2. Natural and Practical Constraints
We have three sets of constraints. First, h[1] contains N practical constraints to ensure that the charging stations are accessible by road and do not lie in geographically infeasible regions. Next,
g[1] contains N[p] natural constraints to ensure that the demand from every population center is fulfilled by charging supply at all the adjacent charger locations. Finally, g[2] contains N natural
constraints to ensure that sufficient charging infrastructure is set up to at least meet the demand.
2.3. Problem Classification
Problem Class: The problem is a constrained nonlinear problem for spatial optimization.
Continuity: The objective function and the constraints are formulated to be continuous functions.
Smoothness: The problem is not smooth even though it is continuous since the objective function and the constraints are summations, therefore their derivatives are not necessarily continuous.
Convexity: The problem is non-convex since the inequality constraints are governed by the Gaussian function, and it is non-convex. However, the objective function and the equality constraints are
Undefined Regions: There are no undefined regions in the problem formulation.
Size: There are 3N variables to solve for, N equality constraints and (N + N[p]) inequality constraints.
2.4. Difficulties in Problem Formulation
Initially the problem was formulated to minimize the sum of the distances between population centers and the charging locations. This raised the possibility of landing at trivial solutions or not
getting any solution at all due to the highly non-convex nature of the spatial optimization.
A new approach was designed to treat the optimization problem as a network flow setup where the demand is absorbed by the charging stations, by also varying the supply at each station. Therefore, the
decision space expanded to include several chargers at each location in addition to the spatial coordinates. The constraints are set up for the demand-supply scenario and willingness to travel to a
public charging station is modeled by a Gaussian distribution.
2.5. Scaling Decisions
Constraint g[2], demand at each population center is fully consumed by the charging stations, was scaled down by a factor of 10 so that the constraints are on the same order of magnitude. The
objective function was also scaled down by the total demand to keep it on the same order of magnitude as the constraints. Additionally, the fitting parameter β was also adjusted to achieve
convergence of the solution.
2.6. Additional Considerations
There are two additional attributes that are not in the model but could be considered in the future. The first is the density of housing types, determined by the American Community Survey [4] . We
could assume that areas with high rates of single-family housing will have a lower demand for public charging stations, as a family that has the space for a personal charger will probably opt for
that option. Additionally, we can add the locations of existing charging stations to make the analysis more robust [6] . This would reduce the total number of stations that are needed to meet demand,
and it would further constrain where the new stations are located.
3. Optimization Study
3.1. Optimization Approach
As mentioned above, we performed our analysis on a “toy problem” of Coronado Island. The island has sixteen population centers located at the centroids of the census block groups. This allowed us to
understand the algorithm and perform a thorough analysis on a much less computationally expensive problem.
For our algorithm, we took a multi-step approach. The first step is to run a genetic algorithm using MATLAB’s “ga” function. We then use the output of the genetic algorithm as starting point for
MATLAB’s “fmincon” function. We chose to use a genetic algorithm to create starting points because we anticipated that our objective function would have many local minima and we needed to introduce
randomness to find a global minimum. We tested different solvers within the “fmincon” function, such as SQP and interior point. After running six trials with each, we found that SQP was not able to
find the global minimum and opted for interior point for the rest of our analysis.
For the two steps above, we kept the constraint that a station is on a road relaxed. We then took the coordinates that were output from MATLAB and input them to ArcGIS, which has a built-in function
that can identify the nearest road vertex. This technically leads to a sub-optimal final solution, but we decided that the computation cost of applying the equality constraint along with the others
far exceeded the benefit.
3.2. Base Case Results
A few initial runs of our solver showed that N = 6 stations was a reasonable value to use as our base case. Because our optimization is partially stochastic, we ran 100 trials to determine the global
minimum. The optimum we found was 1499.94. The optimizers are summarized in Table 2. See Appendix C (Table 8) for more detailed results of this initial study.
Note that we converted the values of x[n] to the number of required chargers based on the scaling factors described in Appendix A. To better visualize the results, we plotted the coordinates in
Figure 2 below. The darker colored diamonds indicate stations with a higher number of chargers.
Table 2. Base case global optimizers.
Figure 2. Solution of charging stations’ location with and without snapping to road.
3.3 Analysis of Results
For a thorough analysis of this base case solution, see Table 3 below.
Further, Table 4 below summarizes the constraint activity for the global minimum. Note that all of the twenty-two inequality constraints were active to some extent.
As shown above, our Coronado Island solution for charging station locations and sizes is intuitive. The locations of the stations are spread in a way where the four larger stations surround the most
densely populated part of the island, and the remaining stations have smaller supply and are located closer to the more sparsely populated parts of the island. Our model assumed that demand would
behave similarly to a population centered Gaussian function. Since the stations in our solution are not of infinite or uniform size, are not co-located, and are not randomly located, we demonstrate
that this assumption holds for this solution. This is best demonstrated by how our solutions spread out. In the case where our g[1] constraint is the only active case, the stations would tend to
cluster in the middle of the population centers. This case would cause the island’s demand to be perfectly supplied, but it would lead to some population centers having more supply than others.
When the g[2] constraint is added each population center cannot be oversupplied, which causes the stations to spread out. In this case, the island would be oversupplied, but each population center
would have a fairly equal supply (see Appendix D). When one looks at Table 4 the Lagrange multipliers for constraint g[2] are larger than g[1]’s indicating that this constraint is more active. This
is desirable because of the equity objective for our problem (equality is guaranteed by constraint g[1]). When our model is run just for the Coronado vs when it is run for the entire county (see
Figure 3) our Coronado station locations change partially due to these effects. In the county-wide case the non-Coronado population centers change how spread out the stations can become, since if
they were to spread out more the stations would be leading to oversupply of some population centers.
4. Sensitivity Analysis
4.1. Constraint Sensitivity
According to Table 4, none of the constraints were truly inactive (i.e. a value of zero), but the level of activity did vary. In general, the value of the Lagrange multipliers for constraint g[1]
were much lower than g[2]. This indicates that the result is much more sensitive to the second constraint than the first.
4.2. Parameterization of Number of Stations
To understand how sensitive our solution is to the number of charging stations (N), we conducted a parametric study. We performed 100 runs for each N = 1 through N = 15, and Table 5 shows a summary
of the results.
Figure 3. Full county preliminary results, snapped to roads (Plotted on GIS).
Table 5. Parameterized Station Count (N)
For N = 1 and N = 2, no minimum was found that satisfied the constraints. This is because the Gaussian w[ij] decays in such a way that outlying populations’ demand cannot be met in any station
placement configuration. For N = 3 through N = 5, a minimum was found that satisfied the constraints, but it was not the global optimum. For N = 6 through N = 15 a global optimum of 1499.94 was
found. For a similar reason as N = 1 and N = 2, the distances to outlying populations w[ij] decay such that x[n] need to be inflated drastically in order to meet demand. The initial parameterization
found an optimum slightly higher for N = 7 through N = 15 (within 0.2). However, when the constraint tolerance was tightened, the higher N values were found to come to the same global optimum, albeit
much slower and in more iterations than for N = 6. The same global optimum for various N gave non-unique placements. As N increased, the {x[lat], x[long]} minimizers remained the same, with some
stations doubling up in placement and occupying the same {x[lat], x[long]}. When this occurred, the co-located stations split the x[n] value, which indicates that the solver was trying to force a
higher N to be equivalent to the global optimum of N = 6.
4.3. Scaling to Full County
We ran the full-county problem once, as shown in Figure 3. We used the optimal N count from the parameterized Coronado Island problem, N = 6, to scale N for the full county. The full county has about
125 times the population, so an N = 750 was used as a scaled N value. However, the ideal N value may not scale linearly. To find the ideal N for the full problem, another parameterization would need
to be conducted. We determined that such an analysis is outside the scope of this project.
The full-county problem contains a few quirks not seen in the Coronado Island case. First, with multiple neighborhoods, the algorithm will attempt to make one station serve different areas separated
by geographic barriers (mountains, ravines, freeways). Much of San Diego’s residential areas are located atop mesas, which means that some stations were allocated to bridges or mountain roads between
such areas; others were snapped to freeways after processing in ArcGIS. Moreover, in the same way that many unsnapped stations in the Coronado Island case were placed in water, one station was placed
in Mexico (we wager it’ll make its way to Zihuatanejo in due time).
Another issue is the underservice of sparse areas. As the width β of the Gaussian functions used in our problem was tuned for the urban Coronado Island, the gradients w[ij] of for many of the
outlying settlements of the county vanish to zero, which prevents any station to move towards fulfilling their demands (or satisfying the constraints in some cases). This can be resolved by tuning β
for the appropriate population density or modulating β to change as a function of local population density to accommodate both dense and sparse areas.
5. Conclusions
The problem we address in this investigation is determining the minimum charging supply needed to equitably satisfy the electric vehicle demand of Coronado Island. These stations are weighted by
their placement relative to the population (demand) centers for Coronado Island. A parametric study of N, the number of stations, was conducted. For this study, we performed 100 iterations for each N
to provide a significant enough sample of our solution space to determine the global solution. The global solution found for Coronado Island was N = 6 stations with a total supply of f = 1499.94. The
placement of the stations in this solution is intuitive. The stations form two clusters, one around the more densely populated northern part of Coronado, and the second cluster around the less
densely populated lower tail of the island. The size of the stations in these clusters also makes sense as the northern cluster has larger stations than the southern cluster. When the sum of the
charging supply is converted to a number of chargers, the solution suggests that roughly twenty-five high-speed chargers would be needed to fulfill the demand for Coronado Island. If we only consider
demand and neglect constraints, only seventeen chargers are needed to meet the demand for Coronado Island; however, our constraints are set up in order to enforce equality and equity of access, both
important factors in building infrastructure networks. Our model reflects a 35% increase in the number of chargers needed, which indicates that our model is non-trivial, and that location is
Our model suggests six optimal locations to place these chargers and therefore has some advantages over a centralized or uniform location scheme. However, like all models, our model makes assumptions
that could affect the real number of chargers needed. For example, we assume that the cost at a charger station is linear with the amount of chargers at each station x[n]. In reality, there are fixed
costs associated with setting up a station. Our model also assumes that the demand comes purely from the residents of the island and neglects people who travel to Coronado during the day and may need
to charge. These factors would cause the numbers of chargers at each station, the location of stations, and the overall number of charging stations to change. Furthermore, we assumed equal demand
among residents. In reality, factors like type of home and commute distance may impact charging demand greatly between residents in a way that was not factored in for this analysis. While these
concerns fell outside the scope of this investigation, they provide a basis for future investigations.
Appendix A. Total Charging Station Estimation
The Scope of our investigation is limited to Coronado Island; however, county-wide demand data was scaled down to estimate charging demand per person for the island of Coronado. Since Coronado Island
is a part of San Diego County, these county-wide figures were then scaled to Coronado’s population. The population density, based on census block groups was obtained from the 2018 census. San Diego
county has a population of 3.34 million [2] , which is unevenly distributed throughout the county. To define the scope of the optimization problem, the objective function should represent the number
of electric vehicles on the road as a function of the population density. More specifically, the number of vehicles that would require charging at any given time was estimated based on some aggregate
assumptions and generalizations about San Diego County that could then get scaled to the known population of Coronado Island. Table 6 shows average distribution of household types with the
availability of home chargers within San Diego region.
First, the housing trends statistics from San Diego’s Regional Planning Agency (SANDAG) shows that the average household size in the region is 2.73 persons per household [5] and each household has 2
cars on average [5] . Therefore, San Diego has approximately 2.45 million on-road vehicles; this is consistent with total estimates of on-road vehicles (2.5 million) [7]
The following assumptions were made to estimate the charging requirements for EVs:
· Average daily usage of each car - 25 miles
· Average maximum range for EVs - 250 miles
· Level 1 chargers - 32 Amp chargers give 25 mi/hr of charging
· Level 2 chargers - 50 Amp chargers give 37 mi/hr of charging
· Level 3 chargers - 300 Amp superchargers give 1000 mi/hr of charging
Optimistically assuming that the charging infrastructure will support level 3 fast charge, the total number of charging stations required in a perfect charging scenario is as follows:
${D}_{\text{miles}}={N}_{\text{cars}}\ast \text{DailyMiles}=61.2\ast {10}^{6}\text{miles}$
Supply per charger:
${S}_{\text{miles}}=1000\text{\hspace{0.17em}}\text{mi/hr}\ast 24\text{\hspace{0.17em}}\text{hr/day}=24,000\text{\hspace{0.17em}}\text{mi/day}$
Now we know the total demand and the amount of demand supplied by each type of charger, so we can determine how many chargers we need assuming no inefficiencies or down-time:
Number of chargers:
Based on the regional average distribution of household types, the availability of home chargers was estimated.
Considering only 50% of the household require access to public chargers on a regular basis and 50% require access to public chargers for opportunity charging. We also factor in 60% charger downtime
due to inefficiencies and day-time/night-time cycle:
Effective Number of chargers:
${N}_{\text{chargers}}=\text{}2550\ast 0.5\ast 1.25\ast 1.6\approx 2550\text{\hspace{0.17em}}\text{chargers}$
This means with the super charger case we would need 2100 chargers to meet the county’s demand for charging. This averages out to approximately 1300 individuals and about 1000 cars using each public
EV charger each day.
When calculated for Coronado is land’s population of 24,697 people, the constant demand model here suggests 17 super charges for the island (ignoring distance to chargers and just meeting baseline
demand). The 1500.08 supply number from our model when re-scaled, leads to a supply of 30,000. As one can see below this means we would need around 23 super chargers on the island of Coronado. Table
7 shows the number of charges supplied to different charging station and scaled according to the number of chargers at each station.
Coronado population to chargers conversion (no account for equity) uniform model:
${#}_{\text{chargers}}=21,400\text{\hspace{0.17em}}\text{people}\ast \frac{1}{300\text{\hspace{0.17em}}\text{people/charger}}=17\text{\hspace{0.17em}}\text{chargers}$
Coronado Supply to chargers conversion our model:
$\begin{array}{c}{#}_{\text{chargers}}=\text{}1500.8\text{\hspace{0.17em}}\text{calculatedsupply}\ast 20\text{\hspace{0.17em}}\text{un-scalingfactor}\ast \frac{1}{1300\text{\hspace{0.17em}}\text
{people/charger}}\\ =23\text{\hspace{0.17em}}\text{chargers}\end{array}$
Table 7. Stations’ supplies scaled to number of chargers.
Appendix B. Matlab Code
Optimization Main Function and Constraints & Variable Functions
function [X_end, F, exitflag, output] = optimization_EV(zip, p_lat, p_lng, demand, N, plots)
%%%%%%%%%%%%%%%%%% Globals %%%%%%%%%%%%%%%%%%%%%%%
% N = 5; % Number of charging stations B = 0.0001; %% fitting parameter
P = [p_lat'; p_lng'; demand']; % Populations centers (lat, long) (j)
%%%%%%%%%%%%% objective function %%%%%%%%%%%%%%%%%% avg = sum(demand) / N;
ga_options = optimoptions(@ga,'NonlinearConstraintAlgorithm','penalty');%, 'InitialPopulationMatrix', pop);
lb_ga = [-117.2*ones(1, N), 32.55*ones(1, N), ones(1, N)];
ub_ga = [-117.1*ones(1, N), 32.75*ones(1, N), avg*N*ones(1, N)]; [X_end, ~] = ga(@(X) objective_ga(X), N*3, [], [], [], [], lb_ga',
ub_ga', @(X) constraints_ga(X, P, B), ga_options);
X_ga = [X_end(1,1:N); X_end(1,(N+1):(N*2)); X_end(1,(2*N+1): (N*3))];
lb = [-117.2*ones(1, N); 32.55*ones(1, N); ones(1, N)];
ub = [-117.1*ones(1, N); 32.75*ones(1, N); avg*N*ones(1, N)]; fmincon_options = optimoptions(@fmincon, 'MaxIterations',
inf, 'MaxFunctionEvaluations', inf, 'Algorithm', 'interior-point'); [X_end, ~, ~, ~] = fmincon(@(X) objective(X, P), X_ga, [], [], [],
[], lb, ub, @(X) constraints(X, P, B), fmincon_options); [X_end, ~, exitflag, output] = fmincon(@(X) objective(X,
P), X_end, [], [], [], [], lb, ub, @(X) constraints(X, P, B), fmincon_options);
F = sum(X_end(3,:));
if plots == 1 figure;
mapshow(zip, 'edgecolor', 'k', 'facecolor', 'none') hold on;
scatter(p_lat, p_lng, 'oc'); if size(X_end, 1) == 3
scatter(X_end(1,:), X_end(2,:), 20,
X_end(3, :), '*', 'linewidth', 1.5);
scatter(X_ga(1,:), X_ga(2,:), '.b', 'linewidth', 1.5); disp(sum(X_end(3, :)));
scatter(X_end(1,1:N), X_end(1,(N+1):(N*2)), 20, X_end(1,
(2*N+1):(N*3)), '*', 'linewidth', 1.5);
function [W, dwdlat, dwdlng] = create_w(X, P, B) % weight function
x_lat = X(1, :);
x_lng = X(2, :);
p_lat = P(1, :);
p_lng = P(2, :);
[x_lat_mesh, p_lat_mesh] = meshgrid(x_lat, p_lat); [x_lng_mesh, p_lng_mesh] = meshgrid(x_lng, p_lng); d_lat = x_lat_mesh - p_lat_mesh;
d_lng = x_lng_mesh - p_lng_mesh; D = d_lat.^2 + d_lng.^2;
W = exp(-D./B);
dwdlat = -2 .* d_lat .* W ./ B;
dwdlng = -2 .* d_lng .* W ./ B;
function [g, h] = constraints(X, P, B) % scaled (Fmincon) x_n_i = X(3, :);
p_o_i = P(3, :);
[W] = create_w(X, P, B);
g1 = ((p_o_i') - sum(x_n_i .*W, 2))/10; g2 = -1 + sum(W, 1)';
g = [g1; g2];
h = [];
function [f, df] = objective(X, P) % scaled (Fmincon) x_n_i = X(3, :);
p_o_i = P(3, :);
f = sum(x_n_i)/sum(p_o_i);
function [g, h] = constraints_ga(X, P, B) %unscaled (GA) Nx = length(X)/3;
X_ga = [X(1, (1):(Nx));X(1, (Nx+1):(Nx*2));X(1, (Nx*2+1):(Nx*3))];
x_n_i = X(1, (Nx*2+1):(Nx*3)); p_o_i = P(3, :);
[W, ~, ~] = create_w(X_ga, P, B);
g1 = (p_o_i' -sum(x_n_i .* W, 2)); g2 = -1 + sum(W, 1)';
g = [g1; g2];
h = [];
function [f, df] = objective_ga(X) % unscaled (GA) Nx = size(X, 2)/3;
x_n_i = X(1, (Nx*2+1):(Nx*3)); f = sum(sum(x_n_i));
Outer Wrapper Function for Running Parame-Terization
%%%%%%%%%%%%%% Population Data %%%%%%%%%%%%%%%%
%%%% Full County
% raw_data = xlsread('toy_data/san_diego_centers.xlsx', 'san_diego_centers'); % full county
% zips = shaperead('Full_data/USA_Counties.shp'); % full county
%%%% Toy Problem
raw_data = xlsread('toy_data/ san_diego_centers.xlsx', 'coronado'); % toy
zips = shaperead('toy_data/2010_Census_5- digit_ZIP_Code_Tabulation_Areas.shp'); % toy
%%%% Processing
zip = zips(1); % creates a polgon p_lng = raw_data(:, 4);
p_lat = raw_data(:, 5); demand = raw_data(:, 6); stats = zeros(200, 16);
%%%% Function Call for N = 1:15
plots = 0; %binary plot or not (1 if all plots) runs = 100;
xs = zeros(300, N+3); for i = 1:runs
p = (i)*3;
[x, n, exitflag, ~] = optimization_EV(zip, p_lat, p_lng, demand, N, plots);
xs([p-2,p-1, p], 4:(N+3)) = x;
xs(p-2, 1) = i;
xs(p-2, 2) = n;
xs(p-2, 3) = exitflag; disp([N,i]);
writematrix(xs,'final_comparison.xls', 'sheet', sprintf('N=
%f', N));
ns = xs(:,2);
ls = ns(ns > 1);
[m] = min(ls);
I = find(ls==m); stat = [N; m; I];
stats(1:size(stat), N-2) = stat;
%%%% Summary Tab
writematrix(stats,'final_comparison.xls', 'sheet', 'stats');
Random Multistart Generator for Inside the Is-Land/County Bounds
function [x, y] = toy_multistart(N, zip) stBB = zip.BoundingBox;
st_minlat = min(stBB(:,2)); st_maxlat = max(stBB(:,2)); st_latspan = st_maxlat - st_minlat; st_minlong = min(stBB(:,1)); st_maxlong = max(stBB(:,1));
st_longspan = st_maxlong - st_minlong; stX = zip.X;
stY = zip.Y;
x = zeros(1, length(st_minlong)); y = zeros(1, length(st_minlat)); for i = 1:N
flagIsIn = 0; while ~flagIsIn
x(i) = st_minlong + rand(1) * st_longspan; y(i) = st_minlat + rand(1) * st_latspan; flagIsIn = inpolygon(x(i), y(i), stX, stY);
Appendix C. Base Case Results: Frequency of Minima
Table 8. Breakdown of 100 runs for N = 6 stations.
Appendix D. Constraint Activity and Spreading
See the Figure 4 below for a visualization of how our constraints affect the spreading behavior of the stations. This explains our results for the toy problem, where most stations spread into the
ocean before being snapped back onto roads.
Figure 4. Constraint values under various geographical spreads. | {"url":"https://www.scirp.org/journal/paperinformation?paperid=130647","timestamp":"2024-11-15T02:48:48Z","content_type":"application/xhtml+xml","content_length":"132114","record_id":"<urn:uuid:cd528757-d065-4dad-85a5-a98705d8a404>","cc-path":"CC-MAIN-2024-46/segments/1730477400050.97/warc/CC-MAIN-20241115021900-20241115051900-00147.warc.gz"} |
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Minimax Theory and its Applications 07 (2022), No. 2, 303--320
Copyright Heldermann Verlag 2022
Normalized Solutions for a System of Fractional Schrödinger Equations with Linear Coupling
Meiqi Liu
Dept. of Mathematical Sciences, Tsinghua University, Beijing, P. R. China
Wenming Zou
Dept. of Mathematical Sciences, Tsinghua University, Beijing, P. R. China
We study the normalized solutions of the following fractional Schr\"odinger system: \begin{equation*} \left\{\ \begin{aligned} &(-\Delta)^s u=\lambda_1 u+\mu_1|u|^{p-2}u+\beta v\quad &\hbox{in}\;\
mathbb{R}^N, \\ &(-\Delta)^s v=\lambda_2 v+\mu_2|v|^{q-2}v+\beta u\quad &\hbox{in}\;\mathbb{R}^N, \end{aligned} \right. \end{equation*} with prescribed mass\ \ $\int_{\mathbb{R}^N} u^2=a$ \ and \ $\
int_{\mathbb{R}^N} v^2=b$,\ \ where $s\in(0,1)$, $2 < p,q \leq2_s^*$, $\beta\in\mathbb{R}$ and $\mu_1,\mu_2,a,b$ are all positive constants. Under different assumptions on $p, q$ and $\beta\in\mathbb
R$, we succeed to prove several existence and nonexistence results about the normalized solutions. Specifically, in the case of mass-subcritical nonlinear terms, we overcome the lack of compactness
by establishing the least energy inequality and obtain the existence of the normalized solutions for any given $a,b > 0$ and $\beta\in\mathbb{R}$. While for the mass-supercritical case, we use the
generalized Pohozaev equality to get the boundedness of the Palais-Smale sequence and obtain the positive normalized solution for any $\beta>0$. Finally, in the fractional Sobolev critical case i.e.,
$p=q=2_s^*$, we give a result about the nonexistence of the positive solution.
Keywords: Fractional Laplacian, Schroedinger system, normalized solutions.
MSC: 35R11, 35B09, 35B33.
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x2 + (y – 3√2x)2 = 1 Solution - Grammar Sikho
x2 + (y – 3√2x)2 = 1 Solution
x2 + (y – 3√2x)2 = 1 Solution. In the realm of mathematics, certain equations stand as fascinating enigmas, captivating the minds of both amateur enthusiasts and seasoned mathematicians alike. One
such equation is x2 + (y – 3√2x)2 = 1. This equation, with its unique structure and properties, holds significant value in the study of algebraic geometry and its applications in various scientific
and engineering disciplines.
Understanding the Components of the Equation x2 + (y – 3√2x)2 = 1
Let’s begin our journey of deciphering this intriguing equation by breaking it down into its fundamental components. At first glance, we can observe two variables, x and y, which represent
coordinates on a two-dimensional plane. The equation, with its combination of square terms, demands thorough scrutiny to comprehend its essence fully.
Graphical Representation
Plotting the given equation on a graph unveils a captivating pattern that holds the key to its understanding. By following a step-by-step guide, we can sketch the graph and interpret its essential
features, such as its shape, symmetry, and intersections with coordinate axes.
Solving the Equation
Equations often serve as puzzles to be solved. The equation x2 + (y – 3√2x)2 = 1 is no exception. We will explore algebraic methods to solve this equation systematically. Each step will be explained
in detail to ensure clarity and comprehension.
Real-Life Applications
Surprisingly, this seemingly abstract equation finds remarkable relevance in practical scenarios. From physics to engineering, we will explore real-life applications that rely on the unique
properties offered by x2 + (y – 3√2x)2 = 1.
Similar Equations and Variations
To expand our understanding, we will investigate equations that share similarities with the given equation. Additionally, exploring variations of the equation will provide valuable insights into the
impact of changes on the graph and solutions.
Historical Significance
Behind every mathematical marvel lies a rich history. Discover the origins of x2 + (y – 3√2x)2 = 1 and the mathematicians who contributed to its development. This historical context sheds light on
its journey through time.
Limitations and Assumptions
No equation is without its limitations. We will discuss the scope and applicability of x2 + (y – 3√2x)2 = 1, as well as the assumptions made during its usage. Understanding these constraints is vital
for accurate interpretations.
Common Mistakes and Troubleshooting
Navigating through the intricacies of this equation might lead to common errors. We will address these mistakes and provide troubleshooting tips to empower readers to tackle challenges effectively.
Also Read:
In conclusion, x2 + (y – 3√2x)2 = 1, though enigmatic, is a remarkable mathematical expression with profound implications. Embracing its intricacies unlocks the doors to a world of possibilities,
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[Solved] Show that (a−b)×(a+b)=2(a×b) - Vectors and 3D Geometry... | Filo
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Consider, [By distributivity of vector product over vector addition]
[Again, by distributivity of vector product over vector addition]
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Question Text Show that
Updated On Nov 6, 2022
Topic Vector Algebra
Subject Mathematics
Class Class 12
Answer Type Text solution:1 Video solution: 1
Upvotes 226
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This is an introduction to quantum computation, a cutting edge field that tries to exploit the exponential power of computers based on quantum mechanics. The course does not assume any prior
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Indicator Capital — Impulsionando os heróis do deep-tech
Apollo 11 (July 16–24, 1969) was the American spaceflight that first landed humans on the Moon. Commander Neil Armstrong and Lunar Module Pilot Buzz Aldrin landed the Apollo Lunar Module Eagle on
July 20, 1969, at 20:17 UTC, and Armstrong became the first person to step onto the Moon's surface six hours and 39 minutes later, on July 21 at 02:56 UTC. Aldrin joined him 19 minutes later, and
they spent about two and a quarter hours together exploring the site they had named Tranquility Base upon landing. Armstrong and Aldrin collected 47.5 pounds (21.5 kg) of lunar material to bring back
to Earth as pilot Michael Collins flew the Command Module Columbia in lunar orbit, and were on the Moon's surface for 21 hours, 36 minutes before lifting off to rejoin Columbia.
Apollo 11 (July 16–24, 1969) was the American spaceflight that first landed humans on the Moon. Commander Neil Armstrong and Lunar Module Pilot Buzz Aldrin landed the Apollo Lunar Module Eagle on
July 20, 1969, at 20:17 UTC, and Armstrong became the first person to step onto the Moon's surface six hours and 39 minutes later, on July 21 at 02:56 UTC. Aldrin joined him 19 minutes later, and
they spent about two and a quarter hours together exploring the site they had named Tranquility Base upon landing. Armstrong and Aldrin collected 47.5 pounds (21.5 kg) of lunar material to bring back
to Earth as pilot Michael Collins flew the Command Module Columbia in lunar orbit, and were on the Moon's surface for 21 hours, 36 minutes before lifting off to rejoin Columbia.
How to play
Use your arrow keys to move the ship and the spacebar to shoot.
boosting deep-tech heroes
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NPTEL Deep Learning - IIT Ropar Week 8 Assignment Answer 2023 » DBC ItanagarNPTEL Deep Learning – IIT Ropar Week 8 Assignment Answer 2023
NPTEL Deep Learning – IIT Ropar Week 8 Assignment Answer 2023
NPTEL Deep Learning – IIT Ropar Week 8 Assignment Solutions
NPTEL Deep Learning – IIT Ropar Week 8 Assignment Answer 2023
1. Which of the following best describes the concept of saturation in deep learning?
Answer :-For Answer Click Here
2. Which of the following methods can help to avoid saturation in deep learning?
Answer :- For Answer Click Here
3. Which of the following is true about the role of unsupervised pre-training in deep learning?
Answer :- For Answer Click Here
4. Which of the following is an advantage of unsupervised pre-training in deep learning?
Answer :- For Answer Click Here
5. What is the main cause of the Dead ReLU problem in deep learning?
Answer :- For Answer Click Here
6. How can you tell if your network is suffering from the Dead ReLU problem?
Answer :- For Answer Click Here
7. What is the mathematical expression for the ReLU activation function?
Answer :- For Answer Click Here
8. What is the main cause of the symmetry breaking problem in deep learning?
Answer :- For Answer Click Here
9. What is the purpose of Batch Normalization in Deep Learning?
Answer :- For Answer Click Here
10. In Batch Normalization, which parameter is learned during training?
Answer :- For Answer Click Here
Course Name Deep Learning – IIT Ropar
Category NPTEL Assignment Answer
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imaginary part of complex numbers, polynomials, or rationals
matrix of real or complex numbers (full or sparse storage), or of polynomials or rationals with real or complex coefficients.
matrix of real numbers, polynomials or rationals, with same sizes than x.
imag(x) is the imaginary part of x. (See %i to enter complex numbers).
c = [ 2 %i, 1+0*%i, 2-3*%i log(-1) (-1)^(1/3) ]
s = sprand(3,3,0.3) + sprand(3,3,0.3)*%i
// Polynomials with complex coefficients:
A = [1-%i*%z (%z-%i)^2]
// Rationals with complex coefficients:
A = [ %z/(1-%z) (1-%z)/%z^2];
B = A(1,[2 1]);
C = A + %i*B
B, imag(C)
See also
• real — real part of complex numbers, polynomials, or rationals
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Episode 22 (Bonus Episode) – Conference Presentation: The Failure of the Einstein Spherical Wave Proof
On June - 30 - 2010
Episode 22 is the Failure of Einstein’s Spherical Wave Proof presentation that I delivered at the 17th Annual NPA Conference held at California State University, Long Beach on 23, June 2010. It is
essentially the “Director’s Cut” of Episode 21, and expands on that material. It shows that Einstein’s Relativity Theory derivation fails because of the failure in the Spherical Wave Proof.
Specifically, this episode covers the following:
• Explains why the Spherical Wave Proof is The Essential Proof that established Relativity Theory
• Shows the failure of Einstein’s Spherical Wave Proof as a failure to develop a second sphere
• Identifies the belief that the proof passes as the result of a “False Positive”, or “Type I Error”
• Discusses implications of the failure on terms like Length Contraction, Space-Time Curvature, and Time Dilation
Viewers who have watched Episode 21 will find much of the material familiar.
[podcast format=”video”]http://www.relativitychallenge.com/media/RelativityChallenge.com-Episode22.m4v[/podcast]
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Electoral reforms and non-transitive dice
You're reading: Features
Electoral reforms and non-transitive dice
Guest post by Andrew, of Manchester MathsJam. Andrew can be found on Twitter as @andrew_taylor and blogs occasionally about maths, among other things, at andrewt.net.
“Grime Dice” are a set of five coloured dice with unusual combinations of numbers on them. The red die, for example, has five fours and a nine. The blue one has three twos and three sevens, so it
loses to the red die about 58% of the time. The green die has five fives and a zero, and will lose to the blue one in 58% of rolls. What makes them interesting is that the green die will beat the red
one in 69% of rolls. These three dice behave rather like rock-paper-scissors — in mathematical terms, they are ‘non-transitive’. The full set of Grime Dice also has a purple and a yellow die, so a
better analogy would be rock-paper-scissors-lizard-Spock.
You might ask which is the best Grime die, and the obvious solution is just to roll all five dice at once, a hundred times, and see which one wins the most times. This is why you should never believe
something simply because it is obvious.
In the 18th Century, the Marquis de Condorcet quite reasonably suggested that if ever there is a single candidate in a general election who would beat each of the other candidates in a head-to-head
ballot, that candidate should be declared the winner. While it’s hard to see how there could ever not be such a candidate, we can construct an electorate that think this way using Grime Dice.
There are 6⁵ = 7,776 possible ways that five dice can land and each of these ways represents a voter. The dice represent ratings of candidates. So this voter:
…rates Labour 9/10, the Conservatives 7/10, the Greens 5/10, the Lib Dems 3/10, and UKIP only 1/10. If you do simply roll all five dice against each other, the blue and yellow dice will each win
about 28% of the rolls, and we’ll end up with a Liberal-Conservative coalition — even though you can see from the first diagram that a majority of voters would rather have Labour than either. But if
Labour were elected, a majority would prefer UKIP, to whom a majority would prefer either the Tories or the Lib Dems. This is Condorcet’s Paradox.
A problem with the “first past the post” system we just used is that even if there was a Condorcet winner, they might not win. Programmers call this a “naïve algorithm”, because it’s simple, easy to
implement and sounds reasonable, but gives results that make no sense. It’s possible for the Condorcet loser to win a first-past-the-post election. You could end up with an MP who is less popular
than any other person on the ballot.
That’s why, in May 2011, we had what the government flattered to call a “referendum” on the Alternative Vote, a rival system more usually known as “Instant Runoff” voting or “what they do on The X
Factor”. After rolling the dice against each other a thousand or so times, we would eliminate the green one, which wins only 8% of the vote, and start again — the idea being that voters can both
safely support the Greens and still have a say in the inevitable two-horse race that will actually decide the result. This time we eliminate UKIP, who have 19% (and predictably didn’t inherit any
second-preference votes from Green supporters). Labour have inherited all the second-preferences so far, and as before the blue and yellow dice tie on 28%. The remaining 46% vote Labour, and
whichever other die you eliminate, red wins.
AV isn’t actually that good either, though. You still won’t always get the Condorcet winner, and while you can’t get the Condorcet loser through the last round, there’s an even stranger quirk I’ll
come to in a moment. So what else is there?
A simple one is “reverse plurality” — a sort of “last past the post” idea, where you vote for the worst candidate and the one with fewest votes wins. In our example, that’s Labour.
Then there’s a “reverse AV” system, called “Coombs’ Method” or “what they do on The Weakest Link”, where you eliminate the candidate with the most of these reverse votes, distribute their votes
between the remaining candidates according to voter preference, and continue until there’s only one left. In our example, the blue and the purple dice each lose 28% of rolls. If we eliminate UKIP,
the Tories crash out in the next round anyway and the Greens win — but if we eliminate the Tories, their inherited votes change things enough that UKIP go on to win the election outright. A
right-wing Tory voter might be better off not voting, so that the Conservatives get knocked out and UKIP win rather than the Greens. This is the strange quirk I mentioned earlier: normal ‘forwards’
AV can punish you for voting in exactly the same way.
There’s a lot of other systems out there, but what I’m trying to put across here is: none of them are very good at all. Worse than that, Arrow’s Impossibility Theorem states that you can’t construct
a system that isn’t at least slightly rubbish — essentially that there may not be a ‘best Grime die’, and while that might sound obvious we should by now know better than to accept it just on that
All voting systems are compromises, but the best I’m aware of is the Schulze method. Also called “beatpath”, it has two of the five coolest names in electoral theory. (The other three are
“Condorcet’s Paradox”, “Marquis de Condorcet” and “Arrow’s Impossibility Theorem”.) This method dispenses with rounds of voting, favourites and least favourites, and runs one big, exhaustive search
over every voter’s full preference list.
A beatpath ballot paper looks exactly like an AV one, and you fill it in in the same way, numbering the candidates in order of preference. The first step in counting the votes is to use these ballots
to build the diagram I posted at the start of this article. The next is to find the best path, along the victory arrows, from each die to each other die. A path is as strong as its weakest link, so
the path from red to yellow is 72%, but there’s also a path from yellow to red (through blue, then green) whose strength is 58%, which would sound like a convincing argument that the yellow die is
better than the red one if you have forgotten the whole point of non-transitive dice. When you’ve finished, and found the strongest path between each pair of dice, you get this:
Red Yellow Blue Purple Green
Red 72% 67% 67% 67%
Yellow 58% 58% 67% 67%
Path from… Blue 67% 67% 67% 67%
Purple 69% 69% 67% 72%
Green 69% 69% 67% 67%
You can use this table to find a nice, sensible, and above all transitive order to put the dice in. The loser is the yellow die, because the path from it to any other die is weaker than the one in
the opposite direction. Because there are only six sides on a die, there are some ties in this election, so depending on how the dice tumble, either the blue or purple die will win, since its path to
any other die is stronger than the reverse path.
Arrow’s theorem still prevents it from being a perfect system. For example, it suffers from the same strange ‘vote punishing’ behaviour as AV, although mostly in strange electorates built from maths
toys where there probably isn’t a right answer, which has to be better than a Parliament full of Condorcet losers.
Another quirk of beatpath is that if you’re a bit cheap and only buy the three-pack of Grime Dice, you still get a blue die, but the green one will beat it. It’s clear why this happens with a model
electorate made out of dice — dice are only “good” or “bad” in comparison to other dice, and these ones are deliberately designed to make that comparison unintuitive. Potential MPs, on the other
hand, can be rated in isolation, and while this won’t stop Condorcet’s Paradox arising from time to time, it does change things slightly: in theory, if you measure strength of feeling, rather than
just ranks, you’re no longer beholden to Arrow’s theorem. A simple method is called “range voting”, where you mark each candidate out of ten, and whoever gets most points altogether wins. In our
example, Labour would win simply because the red die has the most spots on it. The problem now is that this relies on voters honestly reporting mild preferences as such so that the vote counters can
properly ignore them, and that simply isn’t going to happen.
In theory, then, Labour are best, followed by the Lib Dems. But as far as anyone can possibly tell from five-party ballot papers, the Conservatives and UKIP are most popular — and the system we
actually use is even worse.
This is why we are doomed.
2 Responses to “Electoral reforms and non-transitive dice”
1. Peter Rowlett
This post, as featured on BBC Material World 19/04/2012!
The maths of politics. Stand up Mathematician Matt Parker and professor of theoretical physics Andrea Rapisarda look at the role mathematics plays in elections and the way politicians behave.
Andrea argues political decisions would be improved if politicians were selected at random rather than elected, but Matt sees the mathematical flaw in electoral systems, which he likens to
rolling a dice – one where the voters hardly ever get the outcome they wish for. | {"url":"https://aperiodical.com/2012/04/electoral-reforms-and-non-transitive-dice/","timestamp":"2024-11-06T02:30:26Z","content_type":"text/html","content_length":"48034","record_id":"<urn:uuid:e4042ab6-78d6-4e93-9bda-d72c509c903f>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00479.warc.gz"} |
Go to the source code of this file.
subroutine clacp2 (UPLO, M, N, A, LDA, B, LDB)
CLACP2 copies all or part of a real two-dimensional array to a complex array.
Function/Subroutine Documentation
subroutine clacp2 ( character UPLO,
integer M,
integer N,
real, dimension( lda, * ) A,
integer LDA,
complex, dimension( ldb, * ) B,
integer LDB
CLACP2 copies all or part of a real two-dimensional array to a complex array.
Download CLACP2 + dependencies
[TGZ] [ZIP] [TXT]
CLACP2 copies all or part of a real two-dimensional matrix A to a
complex matrix B.
UPLO is CHARACTER*1
Specifies the part of the matrix A to be copied to B.
[in] UPLO = 'U': Upper triangular part
= 'L': Lower triangular part
Otherwise: All of the matrix A
M is INTEGER
[in] M The number of rows of the matrix A. M >= 0.
N is INTEGER
[in] N The number of columns of the matrix A. N >= 0.
A is REAL array, dimension (LDA,N)
The m by n matrix A. If UPLO = 'U', only the upper trapezium
[in] A is accessed; if UPLO = 'L', only the lower trapezium is
LDA is INTEGER
[in] LDA The leading dimension of the array A. LDA >= max(1,M).
B is COMPLEX array, dimension (LDB,N)
[out] B On exit, B = A in the locations specified by UPLO.
[in] LDB LDB is INTEGER
The leading dimension of the array B. LDB >= max(1,M).
Univ. of Tennessee
Univ. of California Berkeley
Univ. of Colorado Denver
NAG Ltd.
September 2012
Definition at line 105 of file clacp2.f. | {"url":"https://netlib.org/lapack/explore-html-3.4.2/dd/d1a/clacp2_8f.html","timestamp":"2024-11-02T01:34:09Z","content_type":"application/xhtml+xml","content_length":"11771","record_id":"<urn:uuid:5b1fc226-01a0-4908-a11a-7c71541e23bb>","cc-path":"CC-MAIN-2024-46/segments/1730477027632.4/warc/CC-MAIN-20241102010035-20241102040035-00209.warc.gz"} |
Introduction to Business Math
For this section you will need the following:
Symbols Used
• [latex]\sum=[/latex] Summation
• [latex]\%C=[/latex] Percent change
• [latex]\text{GAvg}=[/latex] Geometric average
• [latex]n=[/latex] Number of pieces of data
• [latex]\text{SAvg}=[/latex] Simple average
• [latex]w=[/latex] Weight factor for a piece of data
• [latex]\text{WAvg}=[/latex] Weighted average
• [latex]x=[/latex] A piece of data
Formulas Used
• Formula 2.4a – Simple Average
[latex]\begin{align*}\text{SAvg}=\frac{\sum x}{n}\end{align*}[/latex]
• Formula 2.4b – Weighted Average
[latex]\begin{align*}\text{WAvg}=\frac{\sum wx}{\sum w}\end{align*}[/latex]
• Formula 2.4c – Geometric Average
[latex]\begin{align*}\text{GAvg}=\left(\left[\left(1 +\%C_1\right)\times\left(1+\%C_2\right)\times\text{ . . . }\times\left(1+\%C_n\right)\right]^{\frac{1}{n}}-1\right)\times 100\end{align*}[/latex]
No matter where you go or what you do, averages are everywhere. Let’s look at some examples:
• Three-quarters of your student loan is spent. Unfortunately, only half of the first semester has passed, so you resolve to squeeze the most value out of the money that remains. But have you
noticed that many grocery products are difficult to compare in terms of value because they are packaged in different sized containers with different price points?
□ For example, one tube of toothpaste sells in a [latex]125\;\text{mL}[/latex] size for [latex]\$1.99[/latex] while a comparable brand sells for [latex]\$1.89[/latex] for [latex]110\;\text{mL}
[/latex]. Which is the better deal? A fair comparison requires you to calculate the average price per millilitre.
• Your local transit system charges [latex]\$2.25[/latex] for an adult fare, [latex]\$1.75[/latex] for students and seniors, and [latex]\$1.25[/latex] for children. Is this enough information for
you to calculate the average fare, or do you need to know how many riders of each kind there are?
• Five years ago you invested [latex]\$8,000[/latex] in Roller Coasters Inc. The stock value has changed by [latex]9\%[/latex], [latex]−7\%[/latex], [latex]13\%[/latex], [latex]4\%[/latex], and
[latex]−2\%[/latex] over these years, and you wonder what the average annual change is and whether your investment kept up with inflation.
• If you participate in any sport, you have an average of some sort: bowlers have bowling averages; hockey or soccer goalies have a goals against average (GAA); and baseball pitchers have an earned
run average (ERA).
Averages generally fall into three categories. This section explores simple, weighted, and geometric averages.
Simple Averages
An average is a single number that represents the middle of a data set. It is commonly interpreted to mean the “typical value.” Calculating averages facilitates easier comprehension of and comparison
between different data sets, particularly if there is a large amount of data. For example, what if you want to compare year-over-year sales? One approach would involve taking company sales for each
of the [latex]52[/latex] weeks in the current year and comparing these with the sales of all [latex]52[/latex] weeks from last year. This involves [latex]104[/latex] weekly sales figures with [latex]
52[/latex] points of comparison. From this analysis, could you concisely and confidently determine whether sales are up or down? Probably not. An alternative approach involves comparing last year’s
average weekly sales against this year’s average weekly sales. This involves the direct comparison of only two numbers, and the determination of whether sales are up or down is very clear.
In a simple average, all individual data share the same level of importance in determining the typical value. Each individual data point also has the same frequency, meaning that no one piece of data
occurs more frequently than another. Also, the data do not represent a percent change. To calculate a simple average, you require two components:
• The data itself—you need the value for each piece of data.
• The quantity of data—you need to know how many pieces of data are involved (the count), or the total quantity used in the calculation.
[latex]\boxed{2.4\text{a}}[/latex] Simple Average
[latex]\color{blue}{\sum}\color{black}{\text{ is Summation:}}[/latex] This symbol is known as the Greek capital letter sigma. In mathematics it denotes that all values written after it (to the right)
are summed.
[latex]\color{red}{\text{SAvg}}\color{black}{\text{ is Simple Average:}}[/latex] A simple average for a data set in which all data has the same level of importance and the same frequency.
[latex]\color{green}{n}\color{black}{\text{ is Total Quantity:}}[/latex] This is the physical total count of the number of pieces of data or the total quantity being used in the average calculation.
In business, the symbol [latex]n[/latex] is a common standard for representing counts.
[latex]\color{purple}{x}\color{black}{\text{ is Any Piece of Data:}}[/latex] In mathematics this symbol is used to represent an individual piece of data.
As expressed in Formula 2.4a, you calculate a simple average by adding together all of the pieces of data then taking that total and dividing it by the quantity.
Calculate a Simple Average
The steps required to calculate a simple average are as follows:
Step 1: Sum every piece of data.
Step 2: Determine the total quantity involved.
Step 3: Calculate the simple average using Formula 2.4a[latex]\begin{align*}\text{SAvg}=\frac{\sum x}{n}\end{align*}[/latex].
Assume you want to calculate an average on three pieces of data: [latex]95[/latex], [latex]108[/latex], and [latex]97[/latex]. Note that the data are equally important and each appears only once,
thus having the same frequency. You require a simple average.
Step 1: Sum all data:
[latex]\sum x=95+108+97=300[/latex]
Step 2: There are three pieces of data, or [latex]n= 3[/latex].
Step 3: Apply Formula 2.4a[latex]\begin{align*}\text{SAvg}=\frac{\sum x}{n}\end{align*}[/latex]:
The simple average of the data set is [latex]100[/latex].
Although mentioned earlier, it is critical to stress that a simple average is calculated only when all of the following conditions are met:
• All of the data shares the same level of importance toward the calculation.
• All of the data appear the same number of times.
• The data does not represent percent changes or a series of numbers intended to be multiplied with each other.
If any of these three conditions are not met, then either a weighted or geometric average is used depending on which of the above criteria failed. We discuss this later when each average is
It is critical to recognize if you have potentially made any errors in calculating a simple average. Review the following situations and, without making any calculations, determine the best answer.
1) The simple average of [latex]15[/latex], [latex]30[/latex], [latex]40[/latex], and [latex]45[/latex] is:
a. lower than [latex]20[/latex]
b. between [latex]20[/latex] and [latex]40[/latex], inclusive
c. higher than [latex]40[/latex]
The best answer is b. because a simple average should fall in the middle of the data set, which appears spread out between [latex]15[/latex] and [latex]45[/latex], so the middle would be around
2) If the simple average of three pieces of data is [latex]20[/latex], which of the following data do not belong in the data set? Data set: [latex]10[/latex], [latex]20[/latex], [latex]30[/latex],
a. [latex]10[/latex]
b. [latex]20[/latex]
c. [latex]30[/latex]
d. [latex]40[/latex]
The data set that does not belong is d. If the number [latex]40[/latex] is included in any average calculation involving the other numbers, it is impossible to get a low average of [latex]20[/latex].
First quarter sales for Buzz Electronics are as indicated in the table below.
Table 2.4.1
Month 2013 Sales 2014 Sales
January $413,200 $455,875
February $328,987 $334,582
March $359,003 $312,777
Martha needs to prepare a report for the board of directors comparing year-over-year quarterly performance. To do this, she needs you to do the following:
a. Calculate the average sales in the quarter for each year.
b. Express the [latex]2014[/latex] sales as a percentage of the [latex]2013[/latex] sales, rounding your answer to two decimals.
Step 1: What are we looking for?
You need to calculate a simple average, or [latex]SAvg[/latex], for the first quarter in each of [latex]2013[/latex] and [latex]2014[/latex]. Then convert the numbers into a percentage.
Step 2: What do we already know?
You know the three quarters annually:
[latex]\begin{align*}2013:\;\;x_1=\$413,200\;\;x_2=\$328,986\;\;x_3=\$350,003\\[2ex]2014:\;\;x_1=\$455,876\;\; x_2=\$334,582\;\;x_3=\$312,777\end{align*}[/latex]
Additionally, you know that the simple average can be obtained using Formula 2.4a[latex]\begin{align*}\text{SAvg}=\frac{\sum x}{n}\end{align*}[/latex] and that using Formula 3.1b[latex]\begin{align*}
\text{Rate}=\frac{\text{Portion}}{\text{Base}}\end{align*}[/latex] you can calculate [latex]2014[/latex] sales as a percentage of [latex]2013[/latex] sales by treating [latex]2013[/latex] average
sales as the base and [latex]2014[/latex] average sales as the portion.
Step 3: Make substitutions using the information known above.
Calculate simple averages for [latex]2013[/latex] and [latex]2014[/latex] using Formula 2.4a:
[latex]\begin{align*}\text{SAvg}=\frac{\sum x}{n}\end{align*}[/latex]
Simple average for [latex]2013[/latex]:
[latex]\begin{align*}\text{SAvg}_{2013}&=\frac{\$413,200 + \$328,986 + \$350,003}{3}\\[1ex]\text{SAvg}_{2013}&=\frac{\$1,092,189}{3}\\[1ex]\text{SAvg}_{2013}&=\$364,063\end{align*}[/latex]
Simple average for [latex]2014[/latex]:
[latex]\begin{align*}\text{SAvg}_{2014}&=\frac{\$455,876 + \$334,582 + \$312,777}{3}\\[1ex]\text{SAvg}_{2014}&=\frac{\$1,103,235}{3}\\[1ex]\text{SAvg}_{2014}&=\$367,745\end{align*}[/latex]
Finally, apply , substituting [latex]\text{SAvg}_{2013}[/latex] for [latex]\text{Base}[/latex] and [latex]\text{SAvg}_{2014}[/latex] for [latex]\text{Portion}[/latex] and multiplying by [latex]100[/
latex] to obtain percentage. Round the result to two decimal places:
[latex]\begin{align*}\%&=\frac{\text{Portion}}{\text{Base}}\times 100\\[1ex]\%&=\frac{\$367,745}{\$364,063}\times 100\\[1ex]\%&=101.01\%\end{align*}[/latex]
Step 4: Provide the information in a worded statement.
The average monthly sales in [latex]2013[/latex] were [latex]\$364,063[/latex] compared to sales in [latex]2014[/latex] of [latex]\$367,745[/latex]. This means that [latex]2014[/latex] sales are
[latex]101.01\%[/latex] of [latex]2013[/latex] sales.
Weighted Averages
Have you considered how your grade point average (GPA) is calculated? Your business program requires the successful completion of many courses. Your grades in each course combine to determine your
GPA; however, not every course necessarily has the same level of importance as measured by your course credits.
Perhaps your math course takes one hour daily while your communications course is only delivered in one-hour sessions three times per week. Consequently, the college assigns the math course five
credit hours and the communications course three credit hours. If you want an average, these different credit hours mean that the two courses do not share the same level of importance, and therefore
a simple average cannot be calculated.
In a weighted average, not all pieces of data share the same level of importance or they do not occur with the same frequency. The data cannot represent a percent change or a series of numbers
intended to be multiplied with each other. To calculate a weighted average, you require two components:
• The data itself—you need the value for each piece of data.
• The weight of the data—you need to know how important each piece of data is to the average. This is either an assigned value or a reflection of the number of times each piece of data occurs (the
[latex]\boxed{2.4\text{b}}[/latex] Weighted Average
[latex]{\color{red}{\text{WAvg}}}{\color{black}{\text{ is Weighted Average:}}}[/latex] An average for a data set where the data points may not all have the same level of importance or they may occur
at different frequencies.
[latex]\color{blue}{\sum}\color{black}{\text{ is Summation:}}[/latex] This symbol is known as the Greek capital letter sigma. In mathematics it denotes that all values written after it (to the right)
are summed.
[latex]\color{green}{w}\color{black}{\text{ is Weighting Factor:}}[/latex] A number that represents the level of importance for each piece of data in a particular data set. It is either predetermined
or reflective of the frequency for the data.
[latex]\color{purple}{x}\color{black}{\text{ is Any Piece of Data:}}[/latex] In mathematics this symbol is used to represent an individual piece of data.
As expressed in Formula 2.4b, calculate a weighted average by adding the products of the weights and data for the entire data set and then dividing this total by the total of the weights.
Calculate a Weighted Average
The steps required to calculate a weighted average are:
Step 1: Sum every piece of data multiplied by its associated weight.
Step 2: Sum the total weight.
Step 3: Calculate the weighted average using Formula 2.4b[latex]\begin{align*}\text{WAvg}=\frac{\sum wx}{\sum w}\end{align*}[/latex].
Let’s stay with the illustration of the math and communications courses and your GPA. Assume that these are the only two courses you are taking. You finish the math course with an A, translating into
a grade point of [latex]4.0[/latex]. In the communications course, your C+ translates into a [latex]2.5[/latex] grade point. These courses have five and three credit hours, respectively. Since they
are not equally important, you use a weighted average.
Step 1: In the numerator, sum the products of each course’s credit hours (the weight) and your grade point (the data). This means:
[latex]\small(\text{math credit hours}\times\text{math grade point})+(\text{communications credit hours}\times\text{communications grade point})[/latex].
Numerically, this is:
[latex]\begin{align*}\sum wx=\left(5\times 4\right)+\left(3\times 2.5\right)=27.5\end{align*}[/latex]
Step 2: In the denominator, sum the weights. These are the credit hours. You have:
[latex]\begin{align*}\sum w=5+3=8\end{align*}[/latex]
Step 3: Apply Formula 2.4b[latex]\begin{align*}\text{WAvg}=\frac{\sum wx}{\sum w}\end{align*}[/latex] to calculate your GPA.
[latex]\begin{align*}\text{WAvg}&=\frac{\sum wx}{\sum w}\\[1ex]\text{WAvg}&=\frac{27.5}{8}\\[1ex]\text{WAvg}&=3.44\text{ (GPAs have two decimals)}\end{align*}[/latex]
Note that your GPA is higher than if you had just calculated a simple average:
[latex]\begin{align*}\text{SAvg}&=\frac{\sum x}{n}\\[1ex]\text{SAvg}&=\frac{4 + 2.4}{2}\\[1ex]\text{SAvg}&=2.25\end{align*}[/latex]
This happens because your math course, in which you scored a higher grade, was more important in the calculation.
Things To Watch Out For
The most common error in weighted averages is to confuse the data with the weight. If you have the two backwards, your numerator is still correct; however, your denominator is incorrect. To
distinguish the data from the weight, notice that the data forms a part of the question. In the above example, you were looking to calculate your grade point average; therefore, grade point is the
data. The other information, the credit hours, must be the weight.
Paths To Success
The formula used for calculating a simple average is a simplification of the weighted average formula. In a simple average, every piece of data is equally important. Therefore, you assign a value of
1 to the weight for each piece of data. Since any number multiplied by 1 is the same number, the simple average formula omits the weighting in the numerator as it would have produced unnecessary
calculations. In the denominator, the sum of the weights of 1 is no different from counting the total number of pieces of data. In essence, you can use a weighted average formula to solve simple
Determine which information is the data and which is the weight.
3) Rafiki operates a lemonade stand during his garage sale today. He has sold [latex]13[/latex] small drinks for [latex]\$0.50[/latex], [latex]29[/latex] medium drinks for [latex]\$0.90[/latex], and
[latex]21[/latex] large drinks for [latex]\$1.25[/latex]. What is the average price of the lemonade sold?
The price of the drinks is the data, and the number of drinks is the weight.
Determine which information is the data and which is the weight.
4) Natalie received the results of a market research study. In the study, respondents identified how many times per week they purchased a bottle of Coca-Cola. Calculate the average number of
purchases made per week.
Table 2.4.2
Purchases per Week # of People
The purchases per week is the data, and the number of people is the weight.
A mark transcript received by a student at a local college:
Table 2.4.3
Course Grade Credit Hours
Economics 100 B 4
Math 100 A 5
Marketing 100 B+ 3
Communications 100 C 4
Computing 100 A+ 3
Accounting 100 D 4
This chart shows how each grade translates into a grade point:
Table 2.4.4
Grade Grade Point
A+ 4.5
A 4.0
B+ 3.5
B 3.0
C+ 2.5
C 2.0
D 1.0
F 0.0
Calculate the student’s grade point average (GPA). Round your final answer to two decimals.
Step 1: What are we looking for?
The courses do not carry equal weights as they have different credit hours. Therefore, to calculate the GPA you must find a weighted average, or WAvg.
Step 2: What do we already know?
Since the question asked for the grade point average, the grade points for each course are the data, or [latex]x[/latex]. The corresponding credit hours are the weights, or [latex]w[/latex]. This
information can be substituted into Formula 2.4b[latex]\begin{align*}\text{WAvg}=\frac{\sum wx}{\sum w}\end{align*}[/latex] to find the weighted average.
Step 3: Make substitutions using the information known above.
Use the secondary table above to convert each course grade into its grade point:
Table 2.4.5
Course Grade Grade Point Credit Hours
Economics 100 B 3.0 4
Math 100 A 4.0 5
Marketing 100 B+ 3.5 3
Communications 100 C 2.0 4
Computing 100 A+ 4.5 3
Accounting 100 D 1.0 4
Sum every piece of data multiplied by its associated weight:
[latex]\begin{align*}\sum wx&= \sum(\text{Credit Hours}\times\text{Grade Point})\\\sum wx&=(4\times 3.0)+(5\times 4.0)+(3\times 3.5)+(4\times 2.0)+(3\times 4.5)+(4\times 1.0)\\\sum wx&=68\end{align*}
Sum the total weight:
[latex]\begin{align*}\sum w&=4+5+3+4+3+4\\\sum w&=23\end{align*}[/latex]
Substitute into Formula 2.4b:
[latex]\begin{align*}\text{WAvg}&=\frac{\sum wx}{\sum w}\\[1ex]\text{WAvg}&=\frac{68}{23}\\[1ex]\text{WAvg}&=2.96\end{align*}[/latex]
Step 4: Provide the information in a worded statement.
The student’s GPA is [latex]2.96[/latex]. Note that math contributed substantially (almost one-third) to the student’s grade point because this course was weighted heavily and the student performed
Angelika started the month of March owing [latex]\$20,000[/latex] on her home equity line of credit (HELOC). She made a payment of [latex]\$5,000[/latex] on the fifth, borrowed [latex]\$15,000[/
latex] on the nineteenth, and made another payment of [latex]\$5,000[/latex] on the twenty-sixth. Using each day’s closing balance for your calculations, what was the average balance in the HELOC for
the month of March?
Step 1: What are we looking for?
The balance owing in Angelika’s HELOC is not equal across all days in March. Some balances were carried for more days than others. This means you will need to use the weighted average technique and
find [latex]\text{WAvg}[/latex].
Step 2: What do we already know?
You know the following:
Table 2.4.6
Dates Number of Days ([latex]w[/latex]) Balance in HELOC ([latex]x[/latex])
March 1 - March 4 4 $20,000
March 5 - March 18 14 $20,000[latex]-[/latex]$5,000 = $15,000
March 19 - March 25 7 $15,000 + $15,000 = $30,000
March 26 - March 31 6 $30,000[latex]-[/latex]$5,000 = $25,000
Step 3: Make substitutions using the information known above.
Sum every piece of data multiplied by its associated weight:
[latex]\begin{align*}\sum wx&=(4\times \$20,000)+(14\times\$15,000)+(7\times\$30,000)+(6\times\$25,000)\\\sum wx&=\$650,000\end{align*}[/latex]
Sum the total weight:
[latex]\begin{align*}\sum w&=4+14+7+6\\\sum w&=31\end{align*}[/latex]
Calculate the weighted average using Formula 2.4b:
[latex]\begin{align*}\text{WAvg}&=\frac{\sum wx}{\sum w}\\[1ex]\text{WAvg}&=\frac{\$650,000}{31}\\[1ex]\text{WAvg}&=\$20,967.74\end{align*}[/latex]
Step 4: Provide the information in a worded statement.
Over the entire month of March, the average balance owing in the HELOC was [latex]\$20,967.74[/latex]. Note that the balance with the largest weight (March 5 to March 18) and the largest balance
owing (March 19 to March 25) account for almost two-thirds of the calculated average.
Geometric Averages
How do you average a percent change? If sales increase [latex]100\%[/latex] this year and decrease [latex]50\%[/latex] next year, is the average change in sales an increase of
per year? The answer is clearly “no.” If sales last year were [latex]\$100[/latex] and they increased by [latex]100\%[/latex], that results in a [latex]\$100[/latex] increase. The total sales are now
[latex]\$200[/latex]. If sales then decreased by [latex]50\%[/latex], you have a [latex]\$100[/latex] decrease. The total sales are now [latex]\$100[/latex] once again. In other words, you started
with [latex]\$100[/latex] and finished with [latex]\$100[/latex]. That is an average change of nothing, or [latex]0\%[/latex] per year! Notice that the second percent change is, in fact, multiplied
by the result of the first percent change. A geometric average finds the typical value for a set of numbers that are meant to be multiplied together or are exponential in nature.
In business mathematics, you most commonly use a geometric average to average a series of percent changes. Formula 2.4c is specifically written to address this situation.
[latex]\boxed{2.4\text{c}}[/latex] Geometric Average
[latex]{\color{red}{\text{GAvg }}}=\left(\left[\left(1+{\color{blue}{\%C_1}}\right)\times\left(1+{\color{blue}{\%C_2}}\right)\times\text{ . . . }\times\left(1+{\color{blue}{\%C}}_{\color{green}{n}}\
right)\right]^{\frac{1}{\color{green}{n}}}-1\right){\color{purple}{\times 100}}[/latex]
[latex]\color{green}{n}\color{black}{\text{ is Total Quantity:}}[/latex] The physical total count of how many percent changes are involved in the calculation.
[latex]\color{red}{\text{GAvg }}\color{black}{\text{ is Geometric Average:}}[/latex] The average of a series of percent changes expressed in percent format. Every percent change involved in the
calculation requires an additional ([latex]1+\%C[/latex]) to be multiplied under the radical. The formula accommodates as many percent changes as needed.
[latex]\color{purple}{\times 100}\color{black}{\text{ is Percent Conversion:}}[/latex] Because you are averaging percent changes, convert the final result from decimal form into a percentage.
[latex]\color{blue}{\%C}\color{black}{\text{ is Percent Change:}}[/latex] The value of each percent change in the series from which the average is calculated. You need to express the percent changes
in decimal format.
Calculate a Geometric Average
To calculate a geometric average follow these steps:
Step 1: Identify the series of percent changes to be multiplied.
Step 2: Count the total number of percent changes involved in the calculation.
Step 3: Calculate the geometric average using Formula 2.4c[latex]\begin{align*}\text{GAvg}=\left(\left[\left(1 +\%C_1\right)\times\left(1+\%C_2\right)\times\text{...}\times\left(1+\%C_n\right)\right]
^{\frac{1}{n}}-1\right)\times 100\end{align*}[/latex].
Let’s use the sales data presented above, according to which sales increase 100% in the first year and decrease [latex]50\%[/latex] in the second year. What is the average percent change per year?
Step 1: The changes are [latex]\%C_1=+100\%[/latex] and [latex]\%C_2=-50\%[/latex].
Step 2: Two changes are involved, or [latex]n=2[/latex].
Step 3: Apply Formula 2.4c:
[latex]\begin{align*}\text{GAvg}&=\left(\left[\left(1+\%C_1\right)\times\text{ . . . }\times\left(1+\%C_n\right)\right]^{\frac{1}{n}}-1\right)\times 100\\\text{GAvg}&=\left(\left[\left(1+100\%\right)
\times\left(1-50\%\right)\right]^{\frac{1}{2}}-1\right)\times 100\\\text{GAvg}&=\left(\left[2\times 0.50\right]^{\frac{1}{2}}-1\right)\times 100\\\text{GAvg}&=\left(\left[1\right]^{\frac{1}{2}} - 1\
right)\times 100\\\text{GAvg}&=0\%\end{align*}[/latex]
The average percent change per year is [latex]0\%[/latex] because an increase of [latex]100\%[/latex] and a decrease of [latex]50\%[/latex] cancel each other out.
Thing To Watch Out For
A critical requirement of the geometric average formula is that every ([latex]1+\%C[/latex]) expression must result in a number that is positive. This means that the [latex]\%C[/latex] cannot be a
value less than [latex]100\%[/latex] else Formula 2.4c[latex]\begin{align*}\text{GAvg}=\left(\left[\left(1 +\%C_1\right)\times\left(1+\%C_2\right)\times\text{ . . . }\times\left(1+\%C_n\right)\right]
^{\frac{1}{n}}-1\right)\times 100\end{align*}[/latex] cannot be used.
Paths To Success
An interesting characteristic of the geometric average is that it will always produce a number that is either smaller than (closer to zero) or equal to the simple average. In the example, the simple
average of [latex]+100\%[/latex] and [latex]50\%[/latex] is [latex]25\%[/latex], and the geometric average is [latex]0\%[/latex]. This characteristic can be used as an error check when you perform
these types of calculations.
Determine whether you should calculate a simple, weighted, or geometric average.
5) Randall bowled [latex]213[/latex], [latex]245[/latex], and [latex]187[/latex] in his Thursday night bowling league and wants to know his average.
Simple; each item has equal importance and frequency.
Determine whether you should calculate a simple, weighted, or geometric average.
6) Cindy invested in a stock that increased in value annually by [latex]5\%[/latex], [latex]6\%[/latex], [latex]3\%[/latex], and [latex]5\%[/latex]. She wants to know her average increase.
Geometric; these are a series of percent changes on the price of stock.
Determine whether you should calculate a simple, weighted, or geometric average.
7) A retail store sold [latex]150[/latex] bicycles at the regular price of [latex]\$300[/latex] and [latex]50[/latex] bicycles at a sale price of [latex]\$200[/latex]. The manager wants to know the
average selling price.
Weighted; each item has a different frequency.
8) Gonzalez has calculated a simple average of [latex]50\%[/latex] and a geometric average of [latex]60\%[/latex]. He believes his numbers are correct. What do you think?
At least one of the numbers is wrong since a geometric average is always smaller than or equal to the simple average.
From [latex]2006[/latex] to [latex]2010[/latex], WestJet’s year-over-year annual revenues changed by [latex]+21.47\%[/latex], [latex]+19.89\%[/latex], [latex]10.55\%[/latex], and [latex]+14.38\%[/
latex]. This reflects growth from sales of [latex]\$1.751\;\text{billion}[/latex] in [latex]2006[/latex] to [latex]\$2.609\;\text{billion}[/latex] in [latex]2010[/latex].^1 What is the average
percent growth in revenue for WestJet during this time frame?
Step 1: What are we looking for?
Note that these numbers reflect percent changes in revenue. Year-over-year changes are multiplied together, so you would calculate a geometric average, or [latex]\text{GAvg}[/latex].
Step 2: What do we already know?
You know the four percent changes:
You also know that four changes are involved, or [latex]n=4[/latex].
Step 3: Make substitutions using the information known above.
Express the percent changes in decimal format and substitute into Formula 2.4c:
[latex]\begin{align*}\text{GAvg}&=\left(\left[\left(1+\%C_1\right)\times\text{ . . . }\times\left(1+\%C_n\right)\right]^{\frac{1}{n}}-1\right)\times 100\\\text{GAvg}&=\left(\left[\left(1+0.2147\
right)\times\left(1+0.1989\right)\times\left(1-0.1055\right)\times\left(1+0.1438\right)\right]^{\frac{1}{4}}-1\right)\times 100\\\text{GAvg}&=10.483\%\end{align*}[/latex]
Step 4: Provide the information in a worded statement.
On average, WestJet revenues have grown [latex]10.483\%[/latex] each year from [latex]2006[/latex] to [latex]2010[/latex].
Section 2.4 Exercises
Calculate a simple average for questions 1 and 2.
1. [latex]8[/latex], [latex]17[/latex], [latex]6[/latex], [latex]33[/latex], [latex]15[/latex], [latex]12[/latex], [latex]13[/latex], [latex]16[/latex]
2. [latex]\$1,500[/latex], [latex]\$2,000[/latex], [latex]\$1,750[/latex], [latex]\$1,435[/latex], [latex]\$2,210[/latex]
Calculate a weighted average for questions 3 and 4.
3. [latex]4[/latex], [latex]4[/latex], [latex]4[/latex], [latex]4[/latex], [latex]12[/latex], [latex]12[/latex], [latex]12[/latex], [latex]12[/latex], [latex]12[/latex], [latex]12[/latex], [latex]12
[/latex], [latex]15[/latex], [latex]15[/latex]
Table 2.4.7
Data $3,600 $3,300 $3,800 $2,800 $5,800
Weight 2 5 3 6 4
Calculate a geometric average for exercises 5 and 6. Round all percentages to four decimals.
Table 2.4.8
5. [latex]5.4\%[/latex] [latex]8.7\%[/latex] [latex]6.3\%[/latex]
6. [latex]10\%[/latex] [latex]4\%[/latex] [latex]17\%[/latex] [latex]10\%[/latex]
1. [latex]15[/latex]
2. [latex]\$1,779[/latex]
3. [latex]10[/latex]
4. [latex]\$3,795[/latex]
5. [latex]6.7910\%[/latex]
6. [latex]2.6888\%[/latex]
7. If a [latex]298[/latex] mL can of soup costs [latex]\$2.39[/latex], what is the average price per millilitre?
8. Kerry participated in a fundraiser for the Children’s Wish Foundation yesterday. She sold [latex]115[/latex] pins for [latex]\$3[/latex] each, [latex]214[/latex] ribbons for [latex]\$4[/latex]
each, [latex]85[/latex] coffee mugs for [latex]\$7[/latex] each, and [latex]347[/latex] baseball hats for [latex]\$9[/latex] each. Calculate the average amount Kerry raised per item.
9. Stephanie’s mutual funds have had yearly changes of [latex]9.63\%[/latex], [latex]2.45\%[/latex], and [latex]8.5\%[/latex]. Calculate the annual average change in her investment.
10. In determining the hourly wages of its employees, a company uses a weighted system that factors in local, regional, and national competitor wages. Local wages are considered most important and
have been assigned a weight of [latex]5[/latex]. Regional and national wages are not as important and have been assigned weights of [latex]3[/latex] and [latex]2[/latex], respectively. If the
hourly wages for local, regional, and national competitors are [latex]\$16.35[/latex], [latex]\$15.85[/latex], and [latex]\$14.75[/latex], what hourly wage does the company pay?
11. Canadian Tire is having an end-of-season sale on barbecues, and only four floor models remain, priced at [latex]\$299.97[/latex], [latex]\$345.49[/latex], [latex]\$188.88[/latex], and [latex]\
$424.97[/latex]. What is the average price for the barbecues?
12. Calculate the grade point average (GPA) for the following student. Round your answer to two decimals.
Table 2.4.9
Course Grade Credit Hours Grade Grade Point Grade Grade Point
Economics 100 D 5 A+ 4.5 C+ 2.5
Math 100 B 3 A 4.0 C 2.0
Marketing 100 C 4 B+ 3.5 D 1.0
Communications 100 A 2 B 3.0 F 0.0
Computing 100 A+ 3
Accounting 100 B+ 4
13. An accountant needs to report the annual average age (the length of time) of accounts receivable (AR) for her corporation. This requires averaging the monthly AR averages, which are listed below.
Calculate the annual AR average.
Table 2.4.10
Month Monthly AR Average Month Monthly AR Average Month Monthly AR Average
January $45,000 May $145,000 September $185,000
February $70,000 June $180,000 October $93,000
March $85,000 July $260,000 November $60,000
April $97,000 August $230,000 December $50,000
14. From January 2007 to January 2011, the annual rate of inflation has been [latex]2.194\%[/latex], [latex]1.073\%[/latex], [latex]1.858\%[/latex], and [latex]2.346\%[/latex]. Calculate the average
rate of inflation during this period.
7. [latex]\$0.00802/\text{ml}[/latex]
8. [latex]\$6.46[/latex]
9. [latex]5.0821\%[/latex]
10. [latex]\$15.88[/latex]
11. [latex]\$314.83[/latex]
12. [latex]2.74[/latex]
13. [latex]\$125,000[/latex]
14. [latex]1.8666\%[/latex]
Challenge, Critical Thinking, & Other Applications
15. Gabrielle is famous for her trail mix recipe. By weight, the recipe calls for [latex]50\%[/latex] pretzels, [latex]30\%[/latex] Cheerios, and [latex]20\%[/latex] peanuts. She wants to make a
[latex]2[/latex] kg container of her mix. If pretzels cost [latex]\$9.99/\text{kg}[/latex], Cheerios cost [latex]\$6.99/\text{kg}[/latex], and peanuts cost [latex]\$4.95/\text{kg}[/latex], what
is the average cost per [latex]100[/latex] g rounded to four decimals?
16. Caruso is the marketing manager for a local John Deere franchise. He needs to compare his average farm equipment sales against his local Case IH competitor’s sales. In the past three months, his
franchise has sold six [latex]\$375,000[/latex] combines, eighteen [latex]\$210,000[/latex] tractors, and fifteen [latex]\$120,000[/latex] air seeders. His sales force estimates that the Case IH
dealer has sold four [latex]\$320,000[/latex] combines, twenty-four [latex]\$225,000[/latex] tractors, and eleven [latex]\$98,000[/latex] air seeders. Express the Case IH dealer’s average sales
as a percentage of the John Deere dealer’s average sales.
17. You are shopping for shampoo and consider two brands. Pert is sold in a bundle package of two [latex]940[/latex] mL bottles plus a bonus bottle of [latex]400[/latex] mL for [latex]\$13.49[/
latex]. Head & Shoulders is sold in a bulk package of three [latex]470[/latex] mL bottles plus a bonus bottle of [latex]280[/latex] mL for [latex]\$11.29[/latex].
a. Which package offers the best value?
b. If the Head & Shoulders increases its package size to match Pert at the same price per mL, how much money do you save by choosing the lowest priced package?
18. The following are annual net profits (in millions of dollars) over the past four years for three divisions of Randy’s Wholesale:
Cosmetics: [latex]\$4.5[/latex], [latex]\$5.5[/latex], [latex]\$5.65[/latex], [latex]\$5.9[/latex]
Pharmaceutical: [latex]\$15.4[/latex], [latex]\$17.6[/latex], [latex]\$18.5[/latex], [latex]\$19.9[/latex]
Grocery: [latex]\$7.8[/latex], [latex]\$6.7[/latex], [latex]\$9.87[/latex], [latex]\$10.75[/latex]
Rank the three divisions from best performing to worst performing based on average annual percent change.
19. You are shopping for a Nintendo Wii gaming console and visit www.shop.com, which finds online sellers and lists their prices for comparison. Based on the following list, what is the average price
for a gaming console (rounded to two decimals)?
Table 2.4.11
NothingButSoftware.com $274.99
eComElectronics $241.79
NextDayPC $241.00
Ecost.com $249.99
Amazon $169.99
eBay $165.00
Buy.com $199.99
HSN $299.95
Gizmos for Life $252.90
Toys ‘R’ Us $169.99
Best Buy $169.99
The Bay $172.69
Walmart $169.00
20. Juanita receives her investment statement from her financial adviser at Great-West Life. Based on the information below, what is Juanita’s average rate of return on her investments?
Table 2.4.12
Investment Fund Proportion of Entire Portfolio Invested in Fund Fund Rate of Return
Real Estate 0.176 8.5%
Equity Index 0.073 36.2%
Mid Cap Canada 0.100 −1.5%
Canadian Equity 0.169 8.3%
US Equity 0.099 −4.7%
US Mid Cap 0.091 −5.7%
North American Opportunity 0.063 2.5%
American Growth 0.075 −5.8%
Growth Equity 0.085 26.4%
International Equity 0.069 −6.7%
15. [latex]\$0.8082[/latex]
16. [latex]99.0805\%[/latex]
17a. Pert better; Pert=[latex]\$0.005916/\text{ml}[/latex]; H&S=[latex]\$0.006680/\text{ml}[/latex]
17b. Pert saves [latex]\$1.74[/latex]
18. Grocery [latex]11.2853\%[/latex]; Cosmetics [latex]9.4493\%[/latex]; Pharmaceuticals [latex]8.9208\%[/latex]
19. [latex]\$213.64[/latex]
20. [latex]5.9115\%[/latex]
1 WestJet, WestJet Fact Sheet.
THE FOLLOWING LATEX CODE IS FOR FORMULA TOOLTIP ACCESSIBILITY. NEITHER THE CODE NOR THIS MESSAGE WILL DISPLAY IN BROWSER.[latex]\begin{align*}\text{SAvg}=\frac{\sum x}{n}\end{align*}[/latex][latex]\
begin{align*}\text{Rate}=\frac{\text{Portion}}{\text{Base}}\end{align*}[/latex][latex]\begin{align*}\text{WAvg}=\frac{\sum wx}{\sum w}\end{align*}[/latex][latex]\begin{align*}\text{GAvg}=\left(\left
[\left(1 +\%C_1\right)\times\left(1+\%C_2\right)\times\text{ . . . }\times\left(1+\%C_n\right)\right]^{\frac{1}{n}}-1\right)\times 100\end{align*}[/latex]
“3.2: Averages” from Business Math: A Step-by-Step Handbook (2021B) by J. Olivier and Lyryx Learning Inc. through a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License
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100. Elapsed Time | HotDocs Developers
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100. Elapsed Time
Compute elapsed time in days (optional), hours, and minutes.
// Days (omit if you don't care)
SET Days-n TO DAYS FROM( DayStart-n, DayEnd-n )
// Hours
SET Hours-n TO HourEnd-n - HourStart-n
IF Hours-n < 0 // omit section if you aren't calculating days
SET Hours-n TO Hours-n + 24
SET Days-n TO Days-n - 1
// Minutes
SET Minutes-n TO MinEnd-n - MinStart-n
IF Minutes-n < 0
SET Minutes-n TO Minutes-n + 60
SET Hours-n TO Hours-n - 1
IF Hours-n < 0 // omit section if you aren't calculating days
SET Hours-n TO Hours-n + 24
SET Days-n TO Days-n - 1
// Pretty output (optional)
"«Days-n» day"
IF Days-n != 1
RESULT + "s"
RESULT + ", «Hours-n» hour"
IF Hours-n != 1
RESULT + "s"
RESULT + ", and «Minutes-n» minute"
IF Minutes-n != 1
RESULT + "s"
This computation takes a begin time and an end time and calculates the amount of time elapsed between them in hours and minutes. Days are optional. The calculated days, hours and minutes are placed
in number variables, and a nicely formatted string is also returned.
Variables: The computation assumes the following variables:
• DayStart-n (Optional) - A date variable which holds the start date.
• HourStart-n - A number variable which holds the start hour.
• MinStart-n - A number variable which holds the start minute.
• DayEnd-n (Optional) - A date variable which holds the end date.
• HourEnd-n - A number variable which holds the end hour.
• MinEnd-n - A number variable which holds the end minute.
• Days-n (Optional) - A number variable which will be set to the number of elapsed days.
• Hours-n - A number variable which will be set to the number of elapsed hours.
• Minutes-n - A number variable which will be set to the number of elapsed minutes.
Preparing the Variables: Before you can do time calculations, you must make sure that the hours and minutes are placed in separate number variables, and that the hours have been converted into
military time. If your time value is in a text variable, you should use Computation 99: Parsing Time Values. Use Computation #0102: Convert Hours to Military Time if your hours value needs to be
converted to military time.
Days: Elapsed days are optional. To exclude days from the calculation, remove the first section of the computation and the two IF ... END IF blocks marked for removal.
Hours: As long as the hours are in military time, we can arrive at elapsed hours by subtracting HourStart-n from HourEnd-n. If this results in a negative number, we reduce the Days-n variable by one
day and add 24 hours.
Minutes: Like hours, we arrive at elapsed minutes by subtracting MinStart-n from MinEnd-n. If this results in a negative number, we reduce the Hours-n variable by one hour and add 60 minutes to
Caveat: If your start time is later than your end time, you'll get inaccurate results. That should go without saying, but there is always someone ...
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CGPA Calculator - How to Convert CGPA to Percentage?
Colleges and universities worldwide utilize various grading systems. For instance, in India, students receive a percentage determining their division. In contrast, the United States uses a 4-point
GPA (Grade Point Average) system, while Europe employs a 10-point CGPA (Cumulative Grade Point Average) system. The process of convеrting CGPA to a pеrcеntagе involves multiplying your CGPA by 9.5.
It's essential to note that this formula is specific to the CBSE board and may not apply to all еducational institutions. In rеcеnt timеs and gradе points havе bееn usеd to calculatе thе acadеmic
pеrformancе of undеrgraduatе studеnts.
What is CGPA?
The Cumulativе Gradе Point Average (CGPA) is a measure of the student’s performance in an educational program or course. It is the average points scored by a student in all courses or subjects taken
within one semester or academic year.
It is common in most academic institutions to assign grade point values for each of the grades that a student receives, which are typically between 0 and 4 or from 0 to 10. For instance, an “A” would
get a grade point of 4 while the one getting a "B" gets 3. First, all the grade points from courses taken are added up and then divided by how many classes were attended to calculate CGPA.
CGPA to Percentage Formula
Ever scratched your head trying to figure out how your CGPA translates to a percentage? We've all been there - students, parents, and teachers alike. That's why we've put together a simple
CGPA-to-percentage converter. Handcrafted by educators who've faced the same challenge, it's your go-to tool to make sense of those numbers and truly understand your academic journey.
To convert CGPA into percentages, you must multiply your CGPA by 9.5. This is based on the most common scale, where the maximum CGPA is 10. For example, if your CGPA is 9.4, the equivalent percentage
would be 9.4*9.5 = 89.3%.
The Conversion Formula: Percentage = CGPA*9.5
How to Calculate CGPA?
You can calculate CGPA by dividing the total grade points you scored in all major subjects by the total credit points.
CGPA = ∑ (Ci*GPi)/ ∑ Ci
where ,
Ci- Credit Points
GPi - Grade Points
Suppose a student has the following grades and credit values for four courses:
Course Grade Point Credit Value
Mathematics 9 4
English 8 3
Science 7 3
Social Studies 9 2
To calculate the CGPA, we follow these steps:
• Multiply the grade points by the credit values for each course:
1. Mathematics: 9 × 4 = 36
2. English: 8 × 3 = 24
3. Science: 7 × 3 = 21
4. Social Studies: 9 × 2 = 18
• Sum up the products of grade points and credit values : 36 + 24 + 21 + 18 = 99
• Calculate the total credit value: 4 + 3 + 3 + 2 = 12
• Divide the sum of grade points by the total credit value: 99 ÷ 12 = 8.25
Therefore, the student's CGPA is 8.25.
Steps to convert CGPA to percentage
Step 1: First, you must add the marks obtained in all the subjects.
For example , the total grade points are:
1. Subject 1: 8
2. Subject 2: 7
3. Subject 3: 9
4. Subject 4: 7
5. Subject 5: 8
Add the total score obtained: 8+7+9+7+8= 39
• Step 2: Now divide the total by the total number of subjects
For example, CGPA = 39/5 = 7.8
• Step 3 : The CGPA is 7.8
• Step 4: To calculate the percentile, you can multiply the CGPA with 9.5
Percentage= CGPA*9.5
• Step 5: CGPA is 7.8, multiply it with 9.5
Example : 7.8*9.5=74%
This is how you need to calculate your CGPA.
CGPA to Percentage Conversion Chart (9.5 Grading Scale)
CGPA Equivalent Percentage (%) CGPA Equivalent Percentage (%)
10 95 7 66.5
9.9 94.05 6.9 65.55
9.8 93.1 6.8 64.6
9.7 92.15 6.7 63.65
9.6 91.2 6.6 62.7
9.5 90.25 6.5 61.75
9.4 89.3 6.4 60.8
9.3 88.35 6.3 59.85
9.2 87.4 6.2 58.9
9.1 86.45 6.1 57.95
9 85.5 6 57
8.9 84.55 5.9 56.05
8.8 83.6 5.8 55.1
8.7 82.65 5.7 54.15
8.6 81.7 5.6 53.2
8.5 80.75 5.5 52.25
8.4 79.8 5.4 51.3
8.3 78.85 5.3 50.35
8.2 77.9 5.2 49.4
8.1 76.95 5.1 48.45
8 76 5 47.5
7.9 75.05 4.9 46.55
7.8 74.1 4.8 45.6
7.7 73.15 4.7 44.65
7.6 72.2 4.6 43.7
7.5 71.25 4.5 42.75
7.4 70.3 4.4 41.8
7.3 69.35 4.3 40.85
7.2 68.4 4.2 39.9
7.1 67.45 4.1 38.95
– – 4 38
What is CGPA Grading System ?
According to the grading scale, a grade of D is required to pass the board test. So, to pass the exam, you need at least a D or A, B, or C grade. Within a month of the results being released,
students who received grades of E1 or E2 must retake the exam. Refer to the table below to check the grades and grade points for the marks obtained:
Marks Grades Grade Point
91-100 A1 10
81-90 A2 9
71-80 B1 8
61-70 B2 7
51-60 C1 6
41-50 C2 5
33-40 D 4
21-32 E1 –
20 & below E2 –
Why Convert CGPA to Percentage?
The CGPA Grading system lets students and teachers check students’ academic performance and work on their strengths and weaknesses to maximize their results as this factor allows educators to
identify the strengths and weaknesses of students in studies.
Converting CGPA to Percentage is:
• Simple to Use
• Easy to Interpret
• Concise
• More Continuous than letter Grades
• Combined with letter Grades
Benefits of the CGPA System
The CGPA system offers several benefits for both students and educational institutions:
• Reduced Pressure: The CGPA system reduces the pressure of scoring high marks, allowing students to focus on learning without undue stress.
• Advanced Grading Pattern: The grading pattern used in the CGPA system is more advanced, providing a more accurate assessment of a student's performance.
• Identification of Strengths and Weaknesses: The CGPA system helps students identify their strengths and weaknesses, allowing them to work on areas that need improvement.
• Simpler and Continuous Evaluation: The CGPA system provides a more continuous evaluation of a student's performance, making it easier to track progress over time.
Advantages/Disadvantages of the CGPA System
The advantages of the CGPA approach involve combining the various topics' marks and providing a general representation of the applicant's performance. Furthermore, the individual grades allow
students to recognize their strong points and improve on their weaker ones.
Even though CGPA has many advantages, there are a few disadvantages we should be aware of with this grading system.
• Due to the fact that they are simply given a cumulative grade, the level of competition among students is reduced. Given that getting a higher grade is not required, some students could become
• Given that the students who had scores of 90 and 98 received the same grade, the grading system may be regarded unfair.
• The grades are thought to reward a greater percentage than the actual marks scored.
• The scores might not be accurate because those students who scored 89 and 90 are in two different grades, and there is a huge gap between the two. The performance of the students overall may be
CGPA to percentage formula for different Indian Universities
The CGPA grading system is commonly utilized by Indian universities for assessing a student's academic performance throughout the course of a semester as well as the entire program. We've provided
the conversion of CGPA to the percentage for some of the leading universities in India.
CBSE - CGPA to percentage calculation
The CBSE board uses the standard formula of multiplying the CGPA number by 9.5 to convert CGPA to percentage. The student's performance in classes IX and X is determined using this percentage
For example:
If you have CGPA= 9
Percentage is 9*9.5 = 85 %
CBSE revised CGPA to a percentage calculation
Subject-wise percentage calculation = 9.5*GPA of the individual subject
Overall exam percentage calculation = 9.5* CGPA
Anna University CGPA to Percentage Calculator
The Anna University CGPA to percentage calculation applies only to Anna University and its affiliated colleges. To convert your CGPA to a percentage at Anna University, multiply your CGPA by 10.
Percentage = CGPA *10
For instance, if your
CGPA= 7.6
Then your percentage will be CGPA * 10
7.6*10= 76
Referring to the Anna University grades and credit points for semester exams is a good idea.
Marks Grade Credit points
>91 S 10
81-90 A 9
71-80 B 8
61-70 C 7
57-60 D 6
51-56 E 5
<50 U 0
VTU CGPA to Percentage Calculator
The grading system employed by VTU relies on two metrics: SGPA and CGPA. SGPA represents the Semester Grade Point Average and reflects the student's academic performance per semester. To calculate
CGPA, one can employ a formula that involves SGPA.
Visvesvaraya Technological University (VTU) students can follow this formula to calculate their percentage using the CGPA:
CGPA = ∑ (Ci*Si) / ∑ Ci
Ci- Credit Points
Si = SGPA
VTU conversion = (CGPA - 0.75) * 10
Percentage = (Aggregate percentage / 10) + 0.75
KTU CGPA to Percentage Calculator
Kerala Technological University, or APJ Abdul Kalam Technological University, has a 10-point Grade system. The results are given on SGPA and CGPA basis. KTU's calculation technique specifies a
specific formula for calculating CGPA and converting it into a percentage only for KTU students. Check out the KTU CGPA to Percentage Conversion formula.
CGPA = (Ci*GPi)/(Ci)
Ci = Credit granted to a certain course
GPi = Grade Point Average for the Course
Then, after, you can calculate the percentage by multiplying CGPA with 9.5.
GTU CGPA to Percentage Calculator
The Gujarat Technological University offers both two-year and four-year courses, and they use the CGPA and CPT scoring systems. Their students can convert their grades to percentages using a formula
provided by the Board.
Gujarat Technological University (GTU) students can convert their CGPA to a percentage by using the following formula:
Percentage = (CGPA/SPI/CPI - 0.5) * 10)
SPI =Semester Percentage Index
CPI = Cumulative Percentage Index
CGPA = Cumulative Grade Points Average
SPPU CGPA to Percentage Calculator
The SPPU uses a 10-point grading scale. The CGPA and SGPA are used to assess students' performance. For each grade point, a different formula is used to convert CGPA to a percentage SPPU. The college
tries to have a perfect scoring system because the grades earned during the semesters are crucial for future education and employment opportunities. As a result, multiple formulas are utilized for
the conversion.
Grade wise SPPU conversion formula
For Grade D
(4- 4.75 CGPA)
Percentage = (CGPA*6.6)+13.5
For Grade C
(4.75 - 5.25 CGPA)
Percentage = (CGPA*10)-2.5
For Grade B
(5.25 - 5.75 CGPA)
Percentage = (CGPA*10)-2.5
For Grade B+
(5.75 - 6.75 CGPA)
Percentage = (CGPA*5)+26.25
For Grade A
(6.75 - 8.25 CGPA)
Percentage = (CGPA*10)-7.5
For Grade A+
(8.25 - 9.50 CGPA)
Percentage = (CGPA*12)-25
For Grade O
(9.50 - 10.00 CGPA)
Percentage = (CGPA*20)-100
BPUT CGPA to Percentage Calculator
The Biju Patnaik University of Technology uses its own CGPA when calculating percentage scores. Only their college students may use this BPUT CGPA to Percentage conversion method.
percentage = (CGPA-0.50)*10
RGPV CGPA to Percentage Calculator
RGPV applied the credit-based scoring approach. The conversion formula that applies to this university multiplies the grade point average by ten.
Percentage = CGPA * 10
WBUT CGPA to Percentage Calculator
For converting the CGPA to a percentage, WBUT doesn't have a set formula. This assessment system uses a 10-point grading scale. As a result, the formula will be as follows.
Percentage = (CGPA-0.75)*10
MG University CGPA to Percentage Calculator
The basic percentage conversion is followed by the calculation of the CGPA at MG University. Here, to calculate the percentage, the CGPA score will be multiplied by 9.5.
Percentage = CGPA*9.5
SRM University CGPA to Percentage Calculator
The SRM students can use the following table to see the grades they received with respect to the marks they received. Based on the grades received for each individual semester, the overall CGPA will
be calculated. Here, the CGPA ranges from 0 to 10. In order to calculate the percentage equivalent, the CGPA result is generally multiplied by 10.
Percentage= CGPA*10
HNGU CGPA to Percentage Calculator
In HNGU, the calculation for converting CGPA to percentage is as easy as multiplying CGPA by 9.5.
Percentage = CGPA*9.5
Mumbai University CGPA to Percentage Calculator
Mumbai University's conversion mechanism was recently revised in 2018 and now specifies multiple-point scales for different programs. MU has made the following key recommendations:
• A 7-point scale is used for programs in Pure Sciences, Arts, Commerce, and associated subjects. It calculates the student's "actual marks" as a percentage.
• For Engineering, the conversion is accomplished using the 10 Point Grading System, which is as follows:
Percentage = 7.25 * CGPA + 11
MAKAUT CGPA to Percentage Calculator
MAKAUT, also known as Maulana Abul Kalam Azad University of Technology, employs a 7-point grading system that is comparable to the one used by VTU.
Percentage = [CGPA-0.75]* 10
How to Convert CGPA to Percentage for Engineering?
To calculate your CGPA from a given percentage in engineering, divide the percentage by 9.5.
For instance, if you scored 70%, multiply it by 9.5 to get your CGPA, which would be 7.3 out of 10.
Score: 70%
CGPA=7.3 out of 10.
Other Calculator :
SGPA to CGPA
SGPA to Percentage | {"url":"https://www.edmissions.com/blogs/cgpa-to-percentage-calculator","timestamp":"2024-11-01T23:58:17Z","content_type":"text/html","content_length":"219206","record_id":"<urn:uuid:0d1f4a82-67ca-42d4-95d2-748fe12a12e8>","cc-path":"CC-MAIN-2024-46/segments/1730477027599.25/warc/CC-MAIN-20241101215119-20241102005119-00382.warc.gz"} |
Modelling, Dynamics and Control - Chapter four: Classical control design techniques
Chapter four
Classical control design techniques
This chapter is on the theme of linear feedback control. For example, with G(s) representing a system, M(s) a compensator and d an input disturbance signal:
Core skills include:
• How do I analyse the expected behaviour of the closed-loop?
• How do I use analysis tools to facilitate control design?
• Are there classical controller structures which are simple and easy to use?
The focus is on root-loci and frequency response tools alongside lead and lag compensators. It is implicit that students have core competence in some mathematical topics such as polynomials, roots,
complex numbers, exponentials, logarithms, behaviours and Laplace.
Relatively quick overview videos introducing the core topics.
Summary PDF notes:
What are root-loci? How can I use root-loci for analysis and design? What software tools might be useful?
Section two: Frequency response and Bode diagrams
What is frequency response and how do I compute it? What is a Bode diagram and sketching rules for insight? How are Bode diagrams affected by standard compensator structures?
Section three: Nyquist diagrams
What is a Nyquist diagram and how can I use this to assess closed-loop stability? What insights do I gain which lend themselves to control design?
Section four: Gain and phase margins
What are gain and phase margins and why are they important? How do I use these to facilitate systematic analysis and design of expected closed-loop behaviour? How are margins exploited in lead/lag
compensator design?
Section five: Classical feedback analysis tools with MATLAB
Much of control analysis and design requires tedious numerical manipulations which are best handled using computer tools. This section gives an overview of some basic MATLAB tools. | {"url":"https://controleducation.sites.sheffield.ac.uk/chapterclassical","timestamp":"2024-11-13T15:26:19Z","content_type":"text/html","content_length":"120284","record_id":"<urn:uuid:bdb20590-d47f-4468-9d60-08a21a63a3e1>","cc-path":"CC-MAIN-2024-46/segments/1730477028369.36/warc/CC-MAIN-20241113135544-20241113165544-00060.warc.gz"} |
I have credit for AP Statistics. What does this count for? What course can I take next?
Math Placement, Transfer Credit/Credit by Exam, PPL (Math Placement Test) → MATH Credit by Exam & Transfer Credits → I have credit for AP Statistics. What does this count for? What course can I take
4.10. I have credit for AP Statistics. What does this count for? What course can I take next?
See chart below. U of A Math 163 and 263 are not prerequisites to U of A Math or lab science courses (MCB 181L/R, CHEM 141/151, PHYS 102/181) other than Math 302A, which applies to some Education
│[Exam Name │[Required Minimum Score │ [U of A Math Credit ] │ [Notes ] │
│ ] │ ] │ │ │
│Statistics │3 │Math 163, Basic Statistics │Substitutes for Math 106, 107, SBS 200, PSY 230│
│Statistics │4 or 5 │Math 263, Intro to Statistics and Biostatistics│Substitutes for Math 106, 107, SBS 200, PSY 230│
If your major requires Math 112 or higher and you have statistics credit, you may still need to take Math 112 as it is needed as a prerequisite to other courses in your major. Consult your academic | {"url":"https://ua-math-dept.helpspot.com/placement/index.php?pg=kb.page&id=117","timestamp":"2024-11-14T13:39:09Z","content_type":"application/xhtml+xml","content_length":"13477","record_id":"<urn:uuid:6c5d8f98-a09a-42b8-a897-3a04884e9698>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00098.warc.gz"} |
Newtons Method Homework Help By 45+ PhD Qualified Experts
1. Newtons Method Assignment Help
Newtons Method Assignment Help
We all know Isaac newton as one of the greatest scientists of all time. Most people know him because he proposed the law of gravity. But did you know that he was involved in so many scientific
discoveries? There are a number of laws in physics that he came up with, which a reconsidered integral to most physics researches, this has made his name famous. Though he died centuries ago his
theories still exist. Perhaps you would like to know more about the things that he came up with. But here we will focus on one of his mathematical discoveries used to find the root of an equation.
It’s known as the newton’s method or Newton-Rhapson method. It’s named after two scientists who are believed to have contributed to its formulation. There is a lot that you will learn about this
Newton’s method
Newton Rhapson method bears a lot of resemblance to the
secant root finding method
. However, the Newton method is considered superior to other root-finding methods as most practitioners consider it as a more accurate method. Root finding is the process of solving a function by
equating it to zero. The values found are known as the roots of the equation.
Newton’s algorithm operates by finding the first initial guess of the equation. This value is then used to approximate the value of the tangent line. Once we have the tangent line, we then use it to
compute the x-intercept, which will be a better approximation of the roots of the equation. If we do not find the roots, the process can be repeated a number of times until we find the root. The
secret lies in how close the first root is to convergence.
Steps in finding the roots of an equation.
Now that we understand how the Newton Rhapson method works, you can go ahead and find the solution to the equations. Here is a general guideline that you should follow.
1. The first task is to verify if the equation is differentiable. An equation that is not differentiable will not yield the results. Therefore, you cannot go any further with the process of finding
the solution if it’s not differentiable.
2. Once you verified its differential, get its derivative.
3. Guess the initial value to use.
4. Use newton’s iteration formula to get the next value, which is a better approximation of the root.
5. Then iterate the process from step 2 until you find the roots of the equation.
Major problems with newton’s method
Though Newton’s method is considered as the most superior root finding method, it has its challenges, especially when you are computing it. Here are the difficulties associated with Newton’s method.
1. Problems in finding the derivatives
Note that without finding the derivative, there might be no convergence. The effect of this is that we may not find the roots at all. Depending on the equation that we have, we might find it easy to
arrive at the derivative or very complicated. At times, equations do not have a derivative at all. In real life, the analytical expression of the derivative might be extensive.
2. Failure to converge
Newton’s method has its own challenges too. Most of the time, we believe the root-finding method will converge. But it’s not so in most cases. Here are some of the instances when the method might not
• The initial starting point is crucial. If you have a bad starting point, there is a chance that the method might not converge. It might be that the value with which you start does not lie in the
interval where the method converges. In such a case, the bisection method could be the best method. Another issue arises when the starting point is a stationary point. Sometimes the starting
point enters an infinite circle, which prevents the method from converging.
• Derivative issues. Of course, from calculus, we know that a function that is non-differential will not converge. But the effect of this is largely felt if the function used is not differentiable.
3. Stationary points
Sometimes, while calculating the roots of a function, we encounter stationary points. But the stationary points have a zero derivative and we won’t be able to find the roots of the equation.
4. Slow convergence
The process of finding the convergence might be slow, especially where we have a multiplicity of values larger than 1. Again, if the roots are close together, it might need a number of iterations
before the roots are found.
Applications of newton’s method
In its basic form, optimization is finding the maximum and the minimum of a function. In a real-life scenario, you could be asked to find the value that optimizes the profits of the business. This is
quite complex but uses the same idea. Various methods can be used to calculate the optimum values. Newton’s method is one way that can be used to calculate the maximum and minimum values.
Solving the transcendental equation.
These types of equations have transcendental functions. Transcendental functions are functions that cannot be written in polynomial forms. Solving such solutions can be very hard. Examples are log(x)
and sin(x) because they are not polynomials.
Most statistical software is good at finding the roots of an equation using newton’s method. However, most of the time, in your academic life, you will be using Matlab as it has been incorporated
into the academic curriculum of so many institutions.
Generating the roots in Matlab requires you to have good knowledge of creating user-defined functions. It’s not a complex one, and you could make it within no time. In most of the other cases, the
function prompts the user to enter the equation and the first guess. In some cases, you can develop it to solve a specific equation. But in this case, you need not prompt the user to enter any
equation or an initial guess.
Where can I get quality assignment help?
Matlab assignment experts
is a platform dedicated to helping students with their assignments.
Contact us
with that assignment which t you think you are short of time to solve, and we will help you submit it on time along with scoring a high grade. Our skilled
possess in-depth knowledge of newton’s method and can help you solve any task which seems challenging to you. To them, it could just be a piece of cake. They always abide by the instructions issued
with each assignment to ensure you get the grade that you always desire.
For years we have been serving clients from different parts of the world. They have complimented us for the excellent services that we offer. Others have gone to become our brand ambassadors. Giving
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to do next. Once everything is sorted out, we shall start working on the assignment solution. On the other hand, you could click on the ‘submit your assignment’ button on our web page to avail of
assignment help from us. Follow the steps that follow to get us working on the assignment solutions. It generally takes very little time to complete the process.
This sample MatLab assignment solution showcases the derivation of newton’s method using MatLab. It has been used to solve problems in chemical reaction engineering subjects. In this example, two
stirred tank reactors are connected in series. One compound is being converted into another in two reactors. The unconverted compound and the product are being separated in a separation unit and the
unconverted compound is being recycled back. The
states that the total cost of the system is a function of conversion factors in two reactors. The
has demonstrated to find out optimal conversion factors in two reactors for which the total cost of the system is minimum. The cost function in terms of two conversion factors is given as follows,
has derived a version of Newton’s method for solving this problem and has compared it with the solution obtained by fminsearch. He has also calculated the Hessian.
SOLUTION : –
format long
% Function Definition (Enter your Function here):
syms X Y;
f = (X/(Y*(1+X)^2))^0.6+((1-X/Y)/(1-Y)^2)^0.6+6*(1/Y)^0.6;
% Initial Guess (Choose Initial Guesses):
x(1) = 0.2;
y(1) = 0.4;
e = 10^(-8); % Convergence Criteria
i = 1; % Iteration Counter
% Gradient and Hessian Computation:
df_dx = diff(f, X);
df_dy = diff(f, Y);
J = [subs(df_dx,[X,Y], [x(1),y(1)]) subs(df_dy, [X,Y], [x(1),y(1)])]; % Gradient
ddf_ddx = diff(df_dx,X);
ddf_ddy = diff(df_dy,Y);
ddf_dxdy = diff(df_dx,Y);
ddf_ddx_1 = subs(ddf_ddx, [X,Y], [x(1),y(1)]);
ddf_ddy_1 = subs(ddf_ddy, [X,Y], [x(1),y(1)]);
ddf_dxdy_1 = subs(ddf_dxdy, [X,Y], [x(1),y(1)]);
H = [ddf_ddx_1, ddf_dxdy_1; ddf_dxdy_1, ddf_ddy_1]; % Hessian
S = inv(H); % Search Direction
% Optimization Condition:
while norm(J) > e
I = [x(i),y(i)]’;
x(i+1) = I(1)-S(1,:)*J’;
y(i+1) = I(2)-S(2,:)*J’;
i = i+1;
J = [subs(df_dx,[X,Y], [x(i),y(i)]) subs(df_dy, [X,Y], [x(i),y(i)])]; % Updated Jacobian
ddf_ddx_1 = subs(ddf_ddx, [X,Y], [x(i),y(i)]);
ddf_ddy_1 = subs(ddf_ddy, [X,Y], [x(i),y(i)]);
ddf_dxdy_1 = subs(ddf_dxdy, [X,Y], [x(i),y(i)]);
H = [ddf_ddx_1, ddf_dxdy_1; ddf_dxdy_1, ddf_ddy_1]; % Updated Hessian
S = inv(H); % New Search Direction
% Result Table:`
Iter = 1:i;
X_coordinate = x’;
Y_coordinate = y’;
Iterations = Iter’;
T = table(Iterations,X_coordinate,Y_coordinate);
% Plots:
fcontour(f, ‘Fill’, ‘On’);
hold on;
grid on;
% Output:
fprintf(‘Initial Objective Function Value: %d\n\n’,subs(f,[X,Y], [x(1),y(1)]));
if (norm(J) < e)
fprintf(‘Minimum succesfully obtained…\n\n’);
fprintf(‘Number of Iterations for Convergence: %d\n\n’, i);
fprintf(‘Point of Minima: [%d,%d]\n\n’, x(i), y(i));
fprintf(‘Objective Function Minimum Value after Optimization: %f\n\n’, subs(f,[X,Y], [x(i),y(i)]));
% Cost = @(X,Y) (X./(Y.*(1+X)^2)).^0.6+((1-X./Y)/(1-Y).^2)^0.6+6*(1/Y).^0.6;
function b = two_var(v)
x = v(1);
y = v(2);
b = (x./(y.*(1+x)^2)).^0.6+((1-x./y)/(1-y).^2)^0.6+6*(1/y).^0.6; | {"url":"https://www.matlabassignmentexperts.com/newtons-method.html","timestamp":"2024-11-06T21:42:06Z","content_type":"text/html","content_length":"110526","record_id":"<urn:uuid:7e2af793-17e9-44d1-8bd8-0dde3582dfa7>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00711.warc.gz"} |
Convert Decare to Dhur (daa to dhur)
Decare to Dhur Converter
= 0
Decare To Dhur Conversion Table
Unit Conversion Value
1 Decare 2,989.97 Dhur
2 Decare 5,979.95 Dhur
5 Decare 14,949.87 Dhur
10 Decare 29,899.75 Dhur
20 Decare 59,799.49 Dhur
50 Decare 149,498.73 Dhur
100 Decare 298,997.46 Dhur
200 Decare 597,994.92 Dhur
500 Decare 1,494,987.31 Dhur
1000 Decare 2,989,974.62 Dhur
1. What is a decare?
A decare is a metric unit of area equal to 1,000 square meters, commonly used for larger plots of land.
2. What is a dhur?
A dhur is a traditional land measurement unit mainly used in Nepal and India, typically equal to approximately 67.2 square meters.
3. How many dhurs are there in a decare?
One decare is equivalent to about 14.85 dhurs.
4. How do you convert 5 decares to dhurs?
To convert 5 decares to dhurs, multiply by 14.85: 5 Daa x 14.85 = 74.25 Dhur.
5. How many decares are there in 20 dhurs?
To convert 20 dhurs to decares, multiply by 0.0672: 20 Dhur x 0.0672 = 1.344 Daa.
6. Why is it important to convert between these units?
It is vital for accurate land transactions, agricultural planning, and real estate development to ensure clarity regarding land area.
7. Are decares used globally?
Decares are widely used in countries that adhere to the metric system, particularly in Europe and parts of the Middle East.
8. Can the size of a dhur vary?
Yes, while a dhur is commonly recognized as approximately 67.2 square meters, its size can vary slightly depending on regional definitions.
9. How do I measure my land in decares or dhurs?
Land can be measured using surveying tools or by calculating the area based on the lengths of the sides of the land.
10. Where can I find more information on land measurements?
References can be found in agricultural extension service offices, real estate agencies, and various online resources dedicated to land measurement.
About Decare
Decare: Revolutionizing Healthcare Through Technology and Innovation
Healthcare is an essential aspect of human life, and advancements in technology have transformed how we approach healthcare delivery, management, and patient outcomes. One such innovative concept in
this arena is Decare, a term that encompasses various technological solutions and strategies designed to enhance healthcare services. Decare is often associated with a holistic approach to healthcare
management, combining technology, patient-centered care, and data analytics to improve health outcomes and streamline operations.
In this article, we will explore the key components of Decare, its implications for patients and healthcare providers, the technological tools that facilitate its implementation, and its potential
impact on the future of healthcare.
What is Decare?
Decare is not just a single product or service; it represents a comprehensive approach to healthcare that integrates different aspects of health management through digital means. The central idea
behind Decare is to put the patient at the center of the healthcare ecosystem, enabling better communication, more personalized care, and improved access to medical services.
Key Components of Decare
1. Patient-Centric Care: At the core of Decare is the philosophy of patient-centricity, which emphasizes the importance of understanding and addressing the unique needs of each patient. This
involves active participation from patients in their care plans, allowing them to make informed decisions about their health.
2. Integrated Digital Health Solutions: Decare leverages a suite of digital tools, including telemedicine platforms, mobile health applications, and electronic health records (EHRs), to create a
seamless healthcare experience. These tools enable real-time communication between patients and healthcare providers, facilitating timely interventions and support.
3. Data Analytics and AI: The use of data analytics and artificial intelligence (AI) is a cornerstone of Decare. By analyzing vast amounts of health data, healthcare providers can identify trends,
predict outcomes, and optimize treatment plans. AI-powered tools can assist in early detection of diseases and provide personalized recommendations based on individual patient profiles.
4. Preventive Care Focus: Decare promotes a shift from reactive treatments to proactive healthcare through preventive measures. This includes awareness campaigns, regular health screenings, and
lifestyle intervention programs aimed at reducing the incidence of chronic diseases.
5. Interdisciplinary Collaboration: Effective healthcare requires collaboration among various stakeholders, including doctors, nurses, pharmacists, and social workers. Decare fosters
interdisciplinary teamwork by ensuring that all healthcare personnel have access to the same information and can collaborate effectively to coordinate patient care.
Benefits of Decare
The implementation of Decare holds several advantages for both patients and healthcare professionals. Here are some of the most significant benefits:
For Patients:
• Improved Access to Care: Telemedicine features allow patients to consult with healthcare professionals without having to travel to clinics or hospitals, making healthcare more accessible,
especially in rural or underserved areas.
• Enhanced Engagement: Decare encourages patients to engage actively in their healthcare process. With easy access to their health data and treatment plans, patients can make informed choices and
feel empowered regarding their wellness.
• Personalized Treatment Plans: The integration of data analytics enables tailored treatment options based on individual health data, leading to better health outcomes and increased satisfaction.
• Continuous Monitoring: Wearable devices and mobile applications allow for ongoing monitoring of patients’ health metrics, helping to manage chronic conditions effectively and preventing
For Healthcare Providers:
• Streamlined Operations: Electronic health records and integrated systems reduce administrative burdens, allowing healthcare providers to focus on patient care rather than paperwork.
• Better Resource Management: Data-driven insights can aid in the efficient allocation of resources, ensuring that healthcare facilities can respond effectively to patient needs and demands.
• Enhanced Communication: Improved channels of communication among healthcare teams facilitate better coordination of care, leading to fewer errors and improved patient safety.
• Professional Development: Access to ongoing training and development through digital platforms allows healthcare professionals to stay updated with the latest practices and evidence-based
Technological Tools Supporting Decare
Several technological innovations play a pivotal role in the successful implementation of Decare. Here are some of the key tools:
1. Telemedicine Platforms
Telemedicine tools provide necessary virtual consultations, enabling patients to connect with healthcare providers remotely. These platforms offer video conferencing capabilities, secure messaging,
and appointment scheduling, making healthcare more convenient.
2. Mobile Health Applications
Mobile apps can track patient health metrics, remind users of medications, and provide educational resources. Many of these applications integrate with wearable devices to monitor vital signs in
3. Electronic Health Records (EHRs)
EHR systems centralize patient information, allowing healthcare providers to access complete medical histories, lab results, and treatment plans. This improves the continuity of care and ensures
coordinated treatment among different specialists.
4. Artificial Intelligence and Machine Learning
AI algorithms analyze health data to identify patterns that can lead to better predictive analytics. These technologies contribute to early detection of diseases and improve decision-making processes
for healthcare providers.
5. Remote Patient Monitoring Devices
Devices like wearables and home monitoring systems enable continuous tracking of health parameters such as heart rate, blood pressure, and glucose levels. This information can be transmitted to
healthcare providers for real-time assessment and intervention.
The Future of Decare
As healthcare continues to evolve, so too will the principles and practices of Decare. The ongoing advancements in technology promise even greater integration of digital health solutions into
everyday healthcare delivery.
Potential Trends:
• Increased Use of AI and Big Data: We can expect to see even more sophisticated AI applications, which will enhance personalized medicine and predictive healthcare through big data analytics.
• Expansion of Telehealth Services: The growth of telehealth, accelerated by the COVID-19 pandemic, will likely continue, providing greater opportunities for patients to receive care regardless of
• Greater Focus on Mental Health: As mental health becomes an increasingly recognized part of overall health, Decare approaches will incorporate mental health services more prominently within their
• Global Health Initiatives: Decare principles can extend beyond individual patient care to encompass public health initiatives, facilitating broader health promotion and disease prevention
strategies worldwide.
Decare represents a transformative vision for the future of healthcare, where technology, patient involvement, and data analytics converge to create a more efficient, accessible, and personalized
healthcare system. As we continue to innovate and adapt, embracing the tenets of Decare will not only enhance patient experiences but may ultimately revolutionize the way healthcare is delivered
around the globe. Embracing these changes holds the promise of healthier populations and a more sustainable and effective healthcare system for years to come.
About Dhur
Dhur: A Comprehensive Overview
Dhur, also known as Dhura or Dhurva, can refer to different contexts depending on the cultural, geographical, or historical framework in which it is discussed. This article provides a comprehensive
overview of various interpretations of "Dhur," covering its significance in different fields such as linguistics, culture, history, and modern usage.
1. Linguistic Aspects of Dhur
Etymology and Meaning: The term "Dhur" has distinct meanings across different languages. In some contexts, particularly in South Asian languages like Hindi and Urdu, "Dhur" can mean "dust" or
"earth." This connection implies a grounding element, often associated with nature, the environment, and even spirituality.
In Sanskrit, the root "Dhru" refers to something that is steadfast, lasting, or perpetual. This aspect of timelessness connects to various philosophical interpretations in Indian traditions.
Pronunciation and Usage: The pronunciation of "Dhur" may vary slightly by language and regional dialects, but its phonetic structure remains relatively consistent. In conversation, it can be used
metaphorically to signify things that are durable, constant, or reliable.
2. Cultural Significance
Folklore and Myths: In many cultures, dust holds a prominent place in folklore and mythologies, where it symbolizes creation and destruction. For instance, in several indigenous cultures, the earth
or soil is considered sacred, signifying life, fertility, and the cyclical nature of existence.
Dhur in Rituals: Dhur, or dust, often plays a critical role in various rituals. In Hindu traditions, for example, particles of sacred earth or "dharma" are used in ceremonies to invoke blessings. The
sprinkling of soil or dhur during important events symbolizes a connection between the earthly realm and the divine.
3. Historical Context
Dhur in Historical Texts: Historically, the term "Dhur" can be found in ancient texts, particularly those pertaining to geography, astrology, and natural sciences. In Vedic literature, references to
earth and dust appear frequently as metaphors for stability and the foundation of life.
Archaeological Findings: From an archaeological perspective, soil composition and study of ancient civilizations heavily rely on understanding local dhur patterns, especially in regions like the
Indus Valley, Sumer, and Mesoamerica. The dhur in these areas has provided insights into agricultural practices, settlement patterns, and sustainability of ancient societies.
4. Modern Usage
Dhur in Environmental Science: In contemporary environmental science, "Dhur" or dust plays a pivotal role concerning climate change, agriculture, and ecology. Dust storms, for example, can affect air
quality and weather patterns, leading to significant ecological impacts.
In soil science, understanding dhur characteristics—such as texture, salinity, and nutrient content—is vital for effective land management, crop production, and sustainable farming practices. The
study of dhur enables scientists to develop strategies for soil erosion control and restoration of degraded lands.
Urbanization and Dhur: In urban areas, dust pollution is a pressing issue, impacting health and visibility. Policymakers and urban planners must address the management of dust in cities through green
spaces, effective waste management, and public awareness campaigns.
5. Artistic Representations of Dhur
Literature and Poetry: The concept of dhur has inspired countless poets and writers, who use it as a metaphor for resilience, nostalgia, and the passage of time. Dust is often depicted as a reminder
of transience and the inevitability of decay. It serves as a powerful symbol in literature, illustrating both the beauty and harshness of life.
Visual Arts: In visual arts, artists use the imagery of dust and earth to express themes of human struggle, connection to nature, and the passage of time. From abstract paintings that capture the
essence of dust to sculptures made from clay and other earthly materials, dhur permeates creative expression across mediums.
6. Spiritual Connections
Philosophy and Contemplation: In spiritual traditions, dhur is often reflective of the human condition. The acknowledgment of being "dust" serves as a humbling reminder of our mortality and
connection to the earth. Many spiritual leaders emphasize the idea that from dust we come and to dust we shall return, fostering mindfulness and respect for all life forms.
Meditative Practices: Some meditative practices incorporate elements of nature, encouraging participants to connect with the ground beneath them. This practice often invokes senses connected to dhur,
emphasizing the importance of grounding oneself physically and spiritually.
"Dhur" encompasses a rich tapestry of meaning and significance across various domains—from linguistic nuances and cultural traditions to historical significance and modern relevance. Understanding
Dhur requires a multidisciplinary approach that acknowledges its complexity and relevance in both ancient and contemporary contexts.
Whether viewed through the lens of environmental science, spirituality, art, or culture, dhur remains a vital part of our inquiry into the world. It invites us to reflect on our relationship with the
earth, encourages us to consider the impact of our actions, and reminds us of the enduring connections between all living beings. Through this exploration, we recognize dhur not merely as dust, but
as a profound element within the fabric of existence itself.
daadhurDecareDhurdaa to dhurdaa to DhurDecare to DhurDecare to dhurdhur in daadhur in DecareDhur in DecareDhur in daaone daa is equal to how many dhurone Decare is equal to how many Dhurone Decare is
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Convert Decare to DhurConvert Decare to dhur | {"url":"https://www.internettoolwizard.com/convert-units/area/daa/dhur","timestamp":"2024-11-02T22:13:41Z","content_type":"text/html","content_length":"399616","record_id":"<urn:uuid:d3f1f115-3134-4987-8931-2ac75039d2ff>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00291.warc.gz"} |
Dislocation density transients and saturation in irradiated zirconium
AR Warwick and R Thomas and M Boleininger and Ö Kot and G Zilahi and G Ribárik and Z Hegedues and U Lienert and T Ungar and C Race and M Preuss and P Frankel and SL Dudarev, INTERNATIONAL JOURNAL OF
PLASTICITY, 164, 103590 (2023).
DOI: 10.1016/j.ijplas.2023.103590
Zirconium alloys are widely used as the fuel cladding material in pressurized water reactors, accumulating a significant population of defects and dislocations from exposure to neutrons. We present
and interpret synchrotron microbeam X-ray diffraction measurements of proton -irradiated Zircaloy-4, where we identify a transient peak and the subsequent saturation of dislocation density as a
function of exposure. This is explained by direct atomistic simulations showing that the observed variation of dislocation density as a function of dose is a natural result of the evolution of the
dense defect and dislocation microstructure driven by the concurrent generation of defects and their subsequent stress-driven relaxation. In the dynamic equilibrium state of the material developing
in the high dose limit, the defect content distribution of the population of dislocation loops, coexisting with the dislocation network, follows a power law with exponent alpha approximate to 2.2.
This corresponds to the power law exponent of 6 approximate to 3.4 for the distribution of loops as a function of their diameter that compares favourably with the experimentally measured values of 6
in the range 3 <= 6 <= 4.
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Unlock Better Decision-Making with Probability Insights
Ever wondered how you can make more informed decisions in an uncertain world? Probability, often seen as a complex mathematical concept, is actually a powerful tool that can help you navigate through
uncertainty with confidence. In this article, I'll break down the essence of probability and show you how it can be a game-changer in your decision-making process.
Probability is not just about chance; it's about understanding the likelihood of different outcomes and using that knowledge to your advantage. By grasping the fundamentals of probability, you can
assess risks, predict outcomes, and ultimately make better choices in both personal and professional spheres. If you've ever felt overwhelmed by the unpredictability of life, mastering probability
could be the key to unlocking a clearer path forward.
Join me as we delve into the world of probability and discover how this invaluable tool can empower you to make smarter decisions in an increasingly uncertain world. Get ready to embrace a new way of
thinking that will enhance your decision-making skills and set you on a path to success.
The Importance of Probability in Decision-Making
As I dive deeper into the realm of decision-making, I realize the pivotal role that probability plays in shaping our choices. Understanding probability is not merely about luck or chance; it's about
equipping myself with the tools to navigate the intricate web of uncertainties that surround me. Probability, in its essence, empowers me to make well-informed decisions by quantifying the likelihood
of various outcomes.
How Probability Enhances Everyday Choices
In my daily life, whether it's deciding what route to take to work or contemplating whether to carry an umbrella, probability subtly influences my decision-making process. By acknowledging the
probabilistic nature of these choices, I can evaluate the risks involved and make judicious decisions. For instance, when planning a weekend getaway, I can use probability to assess weather forecasts
and choose the destination with the least chance of rain, enhancing my overall experience.
Embracing probability also allows me to optimize resources. For instance, when shopping during a sale, I can calculate the probability of finding the desired item in the right size and color,
enabling me to make efficient purchase decisions. Moreover, in personal finance, understanding the probabilistic nature of investments helps me make strategic choices to grow my assets over time.
Impact on Business and Strategic Planning
In the realm of business, probability serves as a cornerstone for strategic planning and decision-making. I have witnessed firsthand how businesses leverage probability to forecast market trends,
anticipate consumer behavior, and mitigate risks. By analyzing historical data and applying probability models, organizations can make data-driven decisions that optimize performance and drive
Probability also plays a crucial role in risk management within businesses. I recognize that assessing the probability of potential risks allows organizations to implement proactive measures to
mitigate threats and ensure operational continuity. From calculating the likelihood of project delays to estimating revenue fluctuations, probability guides strategic decision-making in navigating
Furthermore, probability acts as a guiding light in the realm of strategic planning. I have observed that by incorporating probability assessments into strategic frameworks, organizations can develop
robust strategies that adapt to changing market dynamics and competitive landscapes. Probability enables businesses to anticipate various scenarios, identify opportunities, and steer their course
towards sustainable growth.
The significance of probability in decision-making cannot be overstated. As I unravel the intricacies of probability, I equip myself with a powerful tool that enhances my decision-making
capabilities, whether in everyday choices or strategic business endeavors. By embracing probability, I navigate uncertainty with confidence, embrace calculated risks, and pave the way for success in
a dynamic and unpredictable world.
Key Concepts in Probability
Understanding Basic Probability Theory
Probability theory is at the core of making decisions in various aspects of life. It is a powerful tool that allows me to quantify uncertainty and make informed choices. By understanding basic
probability concepts, I can analyze the likelihood of different outcomes and make decisions based on rational reasoning rather than intuition.
One fundamental concept in probability theory is the notion of events. An event is an outcome or a set of outcomes of an experiment. I can represent events using sets, with each outcome considered an
element of the set. For example, when rolling a fair six-sided die, the event of rolling a 3 can be represented as {3}.
Another important concept is the sample space, which includes all possible outcomes of an experiment. I can think of the sample space as the universe of all potential results. For the six-sided die
example, the sample space would be {1, 2, 3, 4, 5, 6}.
My understanding of probabilities depends on the concept of probability distributions. A probability distribution describes how the probabilities of different events are spread out. It provides a
clear picture of the likelihood of each possible outcome. I can visualize probability distributions using graphs or tables to make better decisions.
Common Probability Misconceptions
Despite its importance, probability theory is often misunderstood. One common misconception is the belief that past events influence future outcomes. In reality, each event is independent, and past
occurrences do not impact the likelihood of future results. For example, if I flip a coin and it lands on heads ten times in a row, the probability of it landing on heads on the eleventh flip is
still 50%.
Another misconception is the confusion between odds and probability. While both terms refer to the likelihood of an event occurring, they are calculated differently. Probability is the ratio of the
number of favorable outcomes to the total number of outcomes, while odds compare the number of favorable outcomes to the number of unfavorable outcomes. Understanding this distinction is crucial for
accurate decision-making.
I also need to be wary of the gambler's fallacy, which is the mistaken belief that if a certain event occurs more frequently than expected, it is less likely to happen in the future. For instance, if
I roll a fair six-sided die and get a 6 multiple times, the probability of rolling a 6 on the next throw remains the same.
Grasping the key concepts in probability theory and dispelling common misconceptions is essential for making sound decisions in various aspects of life. By applying these principles correctly, I can
navigate uncertainties with confidence and improve my decision-making skills.
Practical Applications of Probability
Case Studies: Probability in Action
Probability plays a crucial role in various real-life scenarios, helping individuals and organizations make informed decisions based on likely outcomes. In a recent study I came across, a
manufacturing company used probability analysis to optimize their production process. By calculating the probability of equipment failure and downtime, they were able to schedule maintenance more
effectively, minimizing disruptions to their workflow. This proactive approach not only saved them time and resources but also ensured smoother operations.
In another case, a retail chain utilized probability to forecast customer demand accurately and adjust their inventory levels accordingly. By tracking sales data and analyzing trends, they could
predict which products were likely to sell well during specific times, preventing stockouts and maximizing profits. This strategic use of probability gave them a competitive edge in the market.
Tools and Techniques for Probability Calculations
When it comes to applying probability in decision-making, there are several tools and techniques available that can help streamline the process and provide more accurate assessments. One such tool is
the probability tree diagram, which visually represents the various outcomes of a decision or event along with their probabilities. I find this tool particularly useful when evaluating multiple
interconnected scenarios and their likelihood of occurrence.
Monte Carlo simulation is another powerful technique that leverages probability to model different possible outcomes of a decision dynamically. By running simulations based on random sampling, I can
assess the potential risks and rewards associated with each choice, allowing for more informed decision-making. This technique is especially valuable in complex scenarios where multiple variables
interact to influence the final outcome.
• Using Bayes' theorem, I can update my beliefs or predictions based on new information that becomes available during the decision-making process. This iterative approach allows me to adjust my
probabilities as I gather more data, leading to more accurate assessments and decisions.
Overall, incorporating these tools and techniques into my decision-making process enhances my ability to assess risks, predict outcomes, and make informed choices based on probability analysis. By
integrating probability into my decision framework, it's not just about taking chances—it's about making calculated and strategic decisions that lead to better outcomes.
Teaching Probability for Better Decision-Making
Educational Methods to Simplify Probability
When teaching probability, I find that it's essential to engage students with real-world examples. By showing how probability affects decisions in everyday life, I can make the subject matter more
relatable and easier to grasp. Explaining concepts like events, sample space, and probability distributions in practical terms helps students connect theory with application. Using probability tree
diagrams is a great way to visually represent these concepts and make them more understandable. Educators can encourage interactive learning by involving students in creating and analyzing these
diagrams for various scenarios.
Role of Probability in Curriculum
Incorporating probability into the curriculum is crucial for developing critical thinking and analytical skills. It allows students to assess risks, evaluate outcomes, and make informed decisions
based on data rather than intuition. Integrating probability into mathematics, statistics, and business courses can provide a strong foundation for students to apply quantitative reasoning in diverse
fields. Including case studies and practical exercises that require probability analysis can further enhance students' ability to think probabilistically and navigate uncertainties effectively.
• Probability offers a systematic approach to decision-making.
• Engaging students with real-world examples enhances learning.
• Integrating probability into the curriculum nurtures analytical skills.
• Practical exercises aid in applying probability concepts effectively.
Understanding probability is a powerful tool that can significantly impact decision-making processes. By quantifying the likelihood of outcomes, we can make more informed choices in various
scenarios. Integrating probability into educational curricula is essential for nurturing critical thinking skills and enhancing analytical abilities. Real-world examples and practical exercises play
a crucial role in making complex probability concepts more accessible and applicable to students. Embracing probability not only empowers individuals to analyze data effectively but also fosters a
systematic approach to decision-making. As we continue to leverage the insights provided by probability, we pave the way for better-informed decisions and a more analytical mindset in navigating the
complexities of today's world.
Frequently Asked Questions
What is the main focus of the article?
The article highlights the importance of probability in decision-making and its practical applications in various areas such as optimizing processes and forecasting demand.
How does the article suggest teaching probability effectively?
The article recommends using real-world examples to make probability concepts relatable and understandable for students, thereby nurturing critical thinking and analytical skills.
What is the significance of integrating probability into the curriculum?
Integrating probability into the curriculum enables students to make informed decisions based on data, fostering analytical skills and enhancing their ability to apply probability concepts
How do practical exercises and case studies contribute to learning probability?
Practical exercises and case studies help enhance students' ability to apply probability concepts effectively, offering a systematic approach to decision-making and fostering analytical skills.
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A uniform disk of radius rolls without slipping with constant angular
A uniform disk of radius rolls without slipping with constant angular velocity ona horizontal surface (see figure below). A point P on the outer edge (rim) of the disk traverses
a path that traces out a cycloid given by the position vector: \vec{r}_{P}=r(\theta-\sin \theta) \hat{\imath}+r(1-\cos \theta) \hat{\jmath} where 's theta reference line is the vertical line shown in
the figure. Compute the velocity Vp of the point P and use it to determine the speed |Vp|| of point P. Your final solution should be of the form: ||Vp| = rwf(8), where ra is the horizontal speed of
the center of mass of the disk and f(theta) is a non-dimensional function of to be determined.
Fig: 1
Fig: 2
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One more than with pictures, numbers and words (1) - More Than or Less Than Maths Worksheets for Year 1 (age 5-6) by URBrainy.com
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Sat preparation permutation combination
sat preparation permutation combination
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JMP: A New Keynesian Model Useful for Policy
Creative Commons CC BY 4.0
I add price-dispersion to a benchmark zero-inflation steady-state New Keynesian model. I do so by assuming the economy has experienced a history of shocks, which have caused the Central Bank to miss
its target for inflation and output, as opposed to the conventional practice of linearizing around a non-stochastic steady state. I then allow the inflation targeting Central Bank to optimize policy.
The results are truly starting.\par The model simultaneously embeds endogenous inflation and interest rate persistence in an institutionally-consistent optimizing framework. This creates a meaningful
trade-off between inflation and output-gap stabilization following demand and technology shocks. This resolves the so-called 'Divine Coincidence', explains the preference for 'coarse-tuning' over
'fine-tuning' and the focus in policy circles on inflation forecast targeting. When estimated the model performs well against a battery of demanding econometric tests. \par Along the way, a novel
econometric test of the 'Divine Coincidence' is developed- it is rejected in favor of a substantial trade-off. A welfare equivalence is derived between a class of New Keynesian models and their
flexible price counterparts suggesting previous proposed resolutions may be inadequate. Finally, a novel paradox relating the 'Divine Coincidence' to 'fine-tuning' stabilization policy is derived. | {"url":"https://tr.overleaf.com/articles/jmp-a-new-keynesian-model-useful-for-policy/dvzbbhpmkqxd","timestamp":"2024-11-14T21:35:52Z","content_type":"text/html","content_length":"205425","record_id":"<urn:uuid:a4cb2f08-85e3-48a1-ba49-19eb6ed16388>","cc-path":"CC-MAIN-2024-46/segments/1730477395538.95/warc/CC-MAIN-20241114194152-20241114224152-00034.warc.gz"} |
Computes a (possibly weighted) contingency table
weighted.table {descriptio} R Documentation
Computes a (possibly weighted) contingency table
Computes a contingency table from one or two vectors, with the possibility of specifying weights.
weighted.table(x, y = NULL, weights = NULL, stat = "freq",
mar = FALSE, na.rm = FALSE, na.value = "NA", digits = 1)
x an object which can be interpreted as factor
y an optional object which can be interpreted as factor
weights numeric vector of weights. If NULL (default), uniform weights (i.e. all equal to 1) are used.
stat character. Whether to compute a contingency table ("freq", default), percentages ("prop"), row percentages ("rprop") or column percentages ("cprop").
mar logical, indicating whether to compute margins. Default is FALSE.
na.rm logical, indicating whether NA values should be silently removed before the computation proceeds. If FALSE (default), an additional level is added to the variables (see na.value argument).
na.value character. Name of the level for NA category. Default is "NA". Only used if na.rm = FALSE.
digits integer indicating the number of decimal places (default is 1)
Returns a contingency table.
Nicolas Robette
See Also
table, assoc.twocat
weighted.table(Movies$Country, Movies$ArtHouse)
version 1.3 | {"url":"https://search.r-project.org/CRAN/refmans/descriptio/html/weighted.table.html","timestamp":"2024-11-05T21:57:06Z","content_type":"text/html","content_length":"3641","record_id":"<urn:uuid:8d7fe626-c076-404d-8af1-b0d345dac292>","cc-path":"CC-MAIN-2024-46/segments/1730477027895.64/warc/CC-MAIN-20241105212423-20241106002423-00782.warc.gz"} |
These units are not compatible
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Hello, I am trying to use this w(k) in my transfer function, but it doesn't work.I can use w(k) separately.Do you have any idea how can I do this?
As Werner says, that evaluation operator at the end of your function definition is the main culprit here.
Removing it cures the problem, and your function should work as you intend, provided fph has unit Hz (as both Werner and Terry said).
And just as an aside, because many new users find using range variables slightly confusing on occasion,
Note the -400 in the first definition; this is to ensure that hk's effective indices start at 0 (add ORIGIN to k-400 if you use a different ORIGIN).
Also, note the vectorized operator in the second definition; this is to ensure there are no unintended side effects from passing a vector to HPWRH. Many a head is scratched trying to work out why
multiplying vectors v & w returns a scalar instead of an element-by-element product (especially when v^2 does return the square of each element of v).
Can you upload the worksheet so the units or absence of units on constants can be checked.
The problem is subtle. The units of the multiplier of "j" must be unitless.
The units of 2*pi in definition of w(k) is in rad (meaning radians) for w of rad/sec.
The units in the denominator of multiplier of j should for the 2*pi as rad and fPH as Hz for rad/sec to cancel out units of w(k)
I am not quite sure what you mean. There is no need to explicitly assign pseudo unit "rad" to any variable. This "unit" is defined as being 1 and has no effect when balancing the units.
As long as f.PH is a quantity with dimension time^-1 (probably Hz or s^-1) all should be OK.I guess that AP_10156885 already had defined f.PH with a correct unit and am pretty sure that the problem
is just the equal sign after the function definition.
The problem is the inline evaluation of your function H...(k).
Delete the equal sign at the end and type H...(k)= in a separate region if you really want to see a list of the 701 values.
I assume that you assigned f.PH a correct unit (dimension time^-1) like 1/s or Hz. | {"url":"https://community.ptc.com/t5/Mathcad/These-units-are-not-compatible/m-p/765258/highlight/true","timestamp":"2024-11-05T07:01:55Z","content_type":"text/html","content_length":"317328","record_id":"<urn:uuid:b3dfe561-04d0-41bd-b882-67a6aafc7cbf>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00435.warc.gz"} |
An Introduction To Statistical Modelling Krzanowski Pdf Free Extra Quality | EZTrade$
An Introduction To Statistical Modelling Krzanowski Pdf Free Extra Quality
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These techniques are important because, despite all their differences, they all involve the same fundamental mathematical operations, and provide insights into the underlying structure of data
(Manly, 2005). The goal of MVAs is to decompose the observed variables into those that are statistically independent and those that are statistically related. Statistical independence is a concept
that is easy to understand; mathematical independence is a more abstract concept. Statistical independence implies that if two variables are linked, then the probability of a value of one (or all) of
the variables is related to a value of the other (or all) variables. This has different meanings for categorical variables (which are discrete) and quantitative variables (which are continuous). For
example, in comparing two values of a categorical variable, the probability of one of those values is related to a value of the other, whereas for quantitative variables, the probability of one of
those values is related to a value of the other. The mathematical (and statistical) relationship between two variables can be represented as a correlation coefficient, which can be positive,
negative, or zero; positive correlation coefficients imply that the variables in the relationship tend to move in the same direction and move together, negative correlation coefficients imply that
the variables in the relationship tend to move in opposite directions and move apart, and zero correlation coefficients imply that the variables in the relationship tend to move together and move in
the same direction, although at different speeds. In addition, there are more elaborate techniques for relating variables that involve the use of two or more correlation coefficients.
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Does God play dice?
For Albert Einstein, the answer is no. But what did he mean? Did the greatest theoretical physicist of all times really miss the bandwagon of quantum physics? What are the real issues of the
controversy that has opposed him to the Copenhagen School (Bohr, Heisenberg …)? Back to the physics of the early twentieth century, its history, philosophy and ideas.
Apples and sun
It all started with a story of apples. Well almost, you might say, do not forget Galilee and Copernicus. You have guessed it, I am not talking about Adam and Eve’s but about the apple that fell on
Newton’s head. In the 17th century, the British scientist laid the foundation for what is now called the “classical mechanics” and managed to model one of the four fundamental interactions: gravity.
His model, fully deterministic, triumphed and appeared then universal for that it predicts the movement of the falling apple as well as the movement of solar system’s stars. This is fabulous! Unify
the phenomena of everyday life with “what goes very high in the sky” in a world that men completely ignored at that time is a historic achievement.
The heyday of Albert
Let’s continue our brief summary of the foundations of modern physics, the journey leading us to the nineteenth century when Maxwell unified within four famous equations the electric and magnetic
forces. This quest of the unification of all physics theories, begun by Maxwelln became even a dream, an ultimate goal for Einstein few decades later. But can we unify electromagnetism (EM) and
All starts from his thoughts on light, and two key findings that hold the attention of Einstein. The first, known since the late nineteenth century, is the fact that light has exactly the same speed
(denoted c) whatever the point of view of the observer: if a passenger in a high-speed train sends a light beam in the train, the velocity of light with respect to the ground is always the same, it
will not be increased by the train’s speed! The problem is that in Maxwell’s equations, light is described as indeed having the famous speed c. But if you look “naively” (using the conventions of …
Newton!) at these equations for a light beam that is already underway in a high-speed train, they show a different speed of light with respect to the ground! In the nineteenth century, many
physicists thought that in fact there was an invisible medium “attached” to light, called Ether, and that only someone who is “attached” to this medium could see the light travelling at c. It is
precisely the theory of relativity by Einstein in the early twentieth century which has given another explanation: the speed of light is the same in all media, but it is necessary, when you look at
the equations of Maxwell in a high-speed train, for example, to be careful not to write them in our daily-life three dimensions, but in four dimensions by adding time. This will keep the same speed
for light whatever the speed of the train in which it is travelling!
On the other hand, but this is this time something Einstein discovered himself in his theory of four dimensional space, the speed of light, this constant we have just mentioned, is the maximum speed
that any phenomenon or object can reach! This calls into question the Newtonian theory of gravity, which stated that the earth is instantly attracted by the Sun, that is to say that the Sun gives
information to the Earth “I’m here, pay attention, I attract you !” with an infinite speed! It is here that Einstein’s relativity, which assumes that the information takes some time to reach the
Earth, provides a theory of gravity which is consistent with Maxwell’s equations now written in the Einstein’s “space-time”.
In a few decades, Einstein challenged the ideas contained in Newtonian mechanics and invented a four-dimensions space-time model that gives coherence together to gravity and electromagnetism, light
being the trigger of the two revolutions. But this theory of relativity does not unify EM and gravity: Einstein’s dream is now to find the master equation for these two major forces that seem to rule
the universe.
But what about quantum physics?
We are in 1930, two new forces have been identified: the weak nuclear interaction, responsible for radioactivity, and the strong nuclear interaction, later used in atomic bombs. They are poorly
theorized, it will be the work of physicists from the standard model and string theory few decades later. But they seem to fit together with Maxwell’s electromagnetism.
The dream of Einstein became the unification of gravitation, which is predominant on a cosmic scale, on the one hand, and on the other hand, the three other forces (EM, weak, strong) prevailing at
the quantum scale. What role played the obsession of the German physicist in his skepticism regarding the recent quantum physics?
While Albert was trying to find a link between gravitation and these three other interactions, quantum physics was already born and developed thanks to Schrödinger, Heisenberg and Bohr among others,
replacing the classical mechanics at the atomic scale. But make no mistake about it, Einstein played a role in the birth of quantum physics, since he used the 1905 model of Planck’s quantum and
defined a particle model for light (previously regarded as a wave) in order to explain the photoelectric effect. This is the counterpart of the wave theory of matter developed by De Broglie. However,
Einstein had many critics, tough constructive, to what quantum theory implied in the probabilistic, philosophical and physics fields. Meanwhile, he continued his work of unification, a work that
would never come along …
“God does not play dice”
This sentence by Einstein gave birth to multiple interpretations, most of which has resulted, and may be an underlying objective, in a discredit of the genius’s post-relativistic work. We will not
fall into this trap. It is very likely that Einstein refused to see Newton’s determinism collapsing, since it remained one of the postulates of its own theory of relativity. How could he accept the
Heisenberg inequality, and especially the fact that the outcome of an experiment can not be estimated without using probabilities? A thought experiment solution would be to say that in fact all
predictable results actually occur, but at a level such that the observer has access only to one of them. Without going into string theory which foresees the existence of seven invisible and folded
on themselves dimensions (like an ant walking on a telephone cable can only see two of three dimensions, length and circularity), Einstein was, after long reflection, seduced by the mathematician
Caluza’s idea of additional spatial dimensions. To overcome this lack of determinism, we can also discuss the concept of “imaginary time” (seen as a pure imaginary complex number) invented by Stephen
Hawking and which allows the realization of all possible trajectories of a particle in the “imaginary time”, as long as one is performed in the “real time”.
Solvay conference (1927): Schrödinger, Bohr, Eisenberg, De Broglie, Curie, Langevin, Dirac, Lorentz, Einstein..
Realism and Positivism
However, we can see this purely as a philosophical quarrel between the realistic and the positivist schools of thought. The realists think we can develop theories that give us an objective knowledge
of the world through the systematic comparison of theory and experiment. Scientific realists, including Einstein and Schrödinger, strongly care for determinism and for objective, independent of the
observer and not hazardous measurement! According to them, it would go against the “common sense” on which physics must still rely. But then what about when Copernicus argued that the Earth was
round, while in the eyes of all it appears of course flat (collective common sense)? This positivist argument was advanced by the so-called Copenhagen School (Bohr, Heisenberg, Jordan, Born …) which
considered quantum physics and physics theories in general as elegant models, results of the imagination of human beings, which should be verified by observations. Where is the difference? Here there
is no immediate sense data (such as “an event occurs in a unique way and not in a hazardous way”), so that it is pointless to ask whether the theory correspond to reality: reality is never
independent of theory!
This positivist way of thinking, although conservative, seems implacable to me. However, some argue that it is outrun by quantum decoherence, which helps to explain mathematically the transition
between “weird” quantum things such as the tunnel effect and the macroscopic world as we see it. In fact, it shows that the reduction of the wave packet due to observation is not in contradiction
with the Schrödinger equation as one might think. I would say that this theory has at least the advantage of “solving” the supposed paradox of Schrödinger’s cat, which is in my opinion a very bad
example of what can explain quantum physics …
Quantum physics, special relativity and measurement
A second problem probably grieved Albert Einstein: that of quantum measurement, which seems to be a projection on a random axis, by the observer, of the observed quantity. So it seems that the
measure is an irregularity in the Schrödinger equation for the wave function, the so-called reduction of the wave packet. One solution to this problem may be to consider that the wave function
represents our knowledge of the system, so that it changes abruptly at the time of measurement. But then, this assumption implies that a measure can reveal something about the system! However,
according to quantum mechanics, only a very large number of measures allow to assess the probability that had the different values to be measured. Thus, all “solutions” quickly seem to fall into the
field of philosophy, which we have already discussed … We have to accept that some variables are not defined before they are measured.
Because he was not convinced, Einstein tried to find a paradox, often called “EPR paradox” in the name of his two acolytes Podolsky and Rosen. Goal: To prove that the states of the particles are
determined at any time. To do this, the thought experiment is to take two entangled photons, ie characterized by the following property: the measure of a quantity related to a photon implies that the
same amount is fully determined for the other. Then we assume that we measure simultaneously (within a time that does not allow exchange of information between the two particles) the position of one
of the photons and the speed of the other. We then obtain a contradiction to the Heisenberg uncertainty principle. Einstein then proposed to enhance quantum physics with the assumption of the
existence of “hidden variables” that are not already included in quantum physics, but are fully deterministic.
However, in the early 80 Alain Aspect gives him wrong with experimental tests. The logic of the EPR paradox is perfect. The assumptions are in fact less. What does the entanglement mean? It can not
be an exchange of information between the two photons, because it does not travel faster than light. In fact, Einstein implicitly assumes that when two photons are sufficiently far apart, we can talk
about the physics characteristics of one specific photon (assuming locality of entanglement). This is directly contradicted by Aspect’s experiments (energy, momentum and polarization measurements),
which showed that the correlation between the two photons is relatively high even if they are far apart: the results of a measurement on a photon depends in a non-local way on the results of the
other photon measurement. The EPR paradox falls through, and the Heisenberg principle is not violated.
Note that nothing here contradicts relativity, which states that the speed of light is a universal upper limit, since there is no way to use the correlations between particles to transmit a signal
faster than the speed: the causality is not violated.
A deeper criticism echoed by Dirac
Finally, there is an other reason which I think by far is the most interesting one, and that made Einstein skeptical. It is the application of another great principle of relativity: E ² = m² c4 + p ²
c ²
Indeed, space and time seem to play the same role in this fundamental equation. But the Schrödinger equation contains a simple time derivative and a double spatial derivative. In fact, Schrödinger
first sought a relativistic equation, now called the Klein-Gordon equation, which contains second derivatives in time and space. The latter being not entirely satisfactory for reasons that we do not
detail here (negative energy solutions), Dirac then derived a system of four equations that contain simple derivatives in time and space. To summarize what we know today, the relativistic Dirac
equations (1928) and the Klein-Gordon (1927) complement, are physically interpretable only in the context of a system of many particles (quantum field theory), and match Schrödinger in the classical
limit. Einstein then had enough to reconcile quantum physics and relativity. Maybe enough to make him change his mind since his “God does not play dice” pronounced in the 1910s.
Ultimately, though some believe that the dispute between Einstein and the Copenhagen school is pure philosophy, it appears that many constructive reasons led Einstein to be skeptical of quantum
mechanics: the role of randomness, shortcomings of the wave function … But keep in mind that unification remained an obsession for him, and that he deliberately sidelined himself from the physics of
atoms. Somehow, we can say that he also refused that a particular philosophy is necessary for the understanding of quantum physics, which can be easily understood by any scientist.
Many philosophers are still working on the issue, whereas it might be time to look at what new physics produce today, and there is much to do. I think about the String Theory, which can not currently
rely on any direct experiment, and lies on the border between philosophy and science. And what about this fierce desire to build a theory of “everything”, a full unification of the laws of nature? Is
it not simply the desire to know the mind of God?
Faced with such questions, I will leave it to the reader to ponder over this quote from Newton, which I very much appreciate: “To myself I am only a child playing on the beach, while vast oceans of
truth lie undiscovered before me.”
1. Very well rounded and balanced article, hitting all the highlights. It is astounding that the concerns that Schrödinger voiced more than 6o years ago with regards to the turn that QM took are
just as relevant today.
To me it always appeared that linear QM must be an approximation that holds in flat space-time for sufficiently isolated systems, but it was just an intuition that seemed sensible to me, not
informed by much actual theoretical analysis, although it directed me to read up on research on the outer edges such as the work of Mendel Sachs, (he made the case that his reformulations of GR
unifies it with EM).
Eventually, I came across Australian physicist Kingsley Jones in a LinkedIn forum, and he published some very interesting work on non-linear deformations of the Schrödinger equation that can
yield a “classic” wave equation that is equivalent to classical Hamilton mechanics. A result that gives very strong theoretical underpinnings to my hunch. (Incidentally Steven Weinberg played
with the same models but missed the connection).
Kingsley is setting out to start a crowd-sourced effort to try to bring QM and QED more in line with Schrödinger’s original vision. I expect this to be a lot of fun and very educational
(irregardless if we will be able to actually deliver on our goal). Contributors with good math skills will be most welcome 🙂
2. Dear Arthur,
Nice survey.
I offer a possibility of reconciling locality with Quantum Mechanics in this suggested mathematical perspective for the EPR argument. | {"url":"https://www.science4all.org/article/does-god-play-dice/","timestamp":"2024-11-03T22:28:27Z","content_type":"text/html","content_length":"66959","record_id":"<urn:uuid:b9156395-0be5-4a96-b739-f95ab59b4b0e>","cc-path":"CC-MAIN-2024-46/segments/1730477027796.35/warc/CC-MAIN-20241103212031-20241104002031-00340.warc.gz"} |
Description This function uses Linear Regression to determine whether a market is moving downward.
Formula Bearish(SERIES, targetbar, LRRange=10, LRType=2)=begin
retval = FALSE
cMin = vchart(LR($1, LRRange, LRType)[targetBar - 1])
pMin = vchart(LR($1, LRRange, LRType)[targetBar])
if cMin < pMin then retval = TRUE
Parameters SERIES
The SERIES directive makes this formula available as a study. You can display this formula in a split chart window.
The index of the bar you want to evaluate. 0 is the current bar, 1 is the first bar back, etc.
The number of bars in the Linear Regression test. The default is 10.
The type of Linear Regression. The default is 2, or Continuous Linear Regression. You can also set this parameter to 3, which gives you a Quadratic Linear Regression.
Return Value TRUE or FALSE
Examples ...
if Bearish($1,targetBar,LRRange,LRType)==TRUE then begin
Comments NA
©2008 Aspen Research Group, Ltd. All rights reserved. Terms of Use. | {"url":"http://www.aspenres.com/Documents/AspenGraphics4.0/Bearish.htm","timestamp":"2024-11-11T18:04:26Z","content_type":"text/html","content_length":"21069","record_id":"<urn:uuid:27a167e9-6eaf-41d1-942a-3396a9374687>","cc-path":"CC-MAIN-2024-46/segments/1730477028235.99/warc/CC-MAIN-20241111155008-20241111185008-00056.warc.gz"} |
Fibonacci Forex Trading Strategy (FREE EA) - Forex James
Fibonacci Forex Trading Strategy (FREE EA)
What Is The Fibonacci Sequence In Forex?
A natural sequence that occurs in nature. “The golden ratio is about 1.618 and is represented by the Greek letter phi, the study of the relationships between numbers, quantities, shapes, and spaces.
relationship between numbers or numerical values.” Learn more about Fibonacci on the web.
Some traders bet their lives on it that they work on the markets as well. In my opinion, they work because the levels coincide with support/resistance levels, where real orders are actually in place.
Some of the popular fib numbers are 38.2 and 61.8, which traders use to enter on a retracement. These are levels in the market where a turning point is anticipated and are regarded as optimal entry
points if you want to participate in the main trend. The 161.8 is also a fib number that traders use to project where the market is headed. It is often used as an exit point or take profit levels.
Let’s Talk About How We Can Combine Fibs With Our Price Action Analysis Below.
Price Structure – Uptrends & Downtrends
Learn to identify swing points and where the impulsive waves are actually at. Impulsive waves are the moves that break previous highs and lows, while corrective waves are the pullback waves. Spotting
HH HL LH LL is one of the basics of price action trading and by understanding it, we can decipher if we’re in an uptrend, downtrend, or a range.
Fibonacci traders believe in the saying The Trend Is Your Friend Until It Bends. We’ve heard this saying many times and many legendary traders have sworn by its truth. After trading for many years, I
mostly trade with the trend and barely make counter-trend trades. It’s just easier if you’re going with the flow i.e. the impulsive wave, as more momentum is present and your TPs can be achieved
If you’re trading with fib retracements, then you’d fall into this category of traders as well.
My advice is that once you’ve identified the long-term trend, don’t go against it.
Fibonacci Retracement In Forex – Pullback Entries
If we were to divide the types of trades into 4 categories, they would be breakout, pullback, counter-trend, and range trades. Breakout traders enter at a worse price, they’d rather wait for a price
confirmation for more assurance. Counter-trend traders go against the herd, hoping for a quick pullback to make quick profits, while the main trend makes a pullback. Range traders go back and forth
between the identified ranges and do not trade outside of them when a trend happens.
Fibonaccis are for pullback traders, who wish to enter the markets on a discount, i.e. a better price. They’ve identified a level in which they’d be interested to enter, in hopes that the trend will
continue. Pullback traders get hurt at the end of the trend when it falls into a range or turns in a different direction. If they’re good enough, they may still get out of a smaller loss or break
Forex Fibonacci Extension
The other benefit of fibs is using it to measure our exit target/ TP. By pulling the fibs on the corrective wave or the BC leg (in the ABCD pattern), we can measure where the price is likely headed
to. For me, the 161.8% level has been pretty accurate and if you’re still uncertain of where you can place your TP, you may use the fib extension. At times when the trend is really strong, we can
even reach the 261.8% level, but usually not before a period of range, after the 161.8% level has been hit. Again, use fibs with a slight discretion and keep in mind that the lines should be regarded
as zones, rather than just a line.
If you prefer using the ABCD pattern to measure the target, that would work as well. Whether you’re using the fibs or the pattern, these are simply just tools to measure the projection of price. As
long as you’re making trades that can yield a bigger reward-to-risk ratio, you’ll be fine in the long run.
Read more about the Parabolic SAR EA here
Read more about the Trendline Break EA here
How The Fibonacci EA Works
• Change the name of your Fibonacci to just “Fibo”.
• Enter the desired mode – Choose BUY or SELL. The EA can only be in 1 mode at a time. For buys, it will enter on price dips and for sells it will look for rallies.
• SL – Placed below or above the high or low of the last bars.
• TP – A multiple of the distance between the entry and stop loss distance. An easy way to measure risk to reward.
• MagicNB – There can only be one open trade with the same magicNB. If you choose to add the Fibo EA to a second chart, be sure to use a different number.
Slippage, SL padding, and the Last Stop loss bars are recommended to be left at their default settings.
Drawing the fibs is pretty straightforward, you can watch the video below for an easier demonstration and also more details on how to use the EA.
The Trigger: Fibonacci Forex Trading Strategy
Depending on the level of retracement you choose, for buys, we wait for a candle close below it and the next candle to close above the line. For sales, we wait for a candle to close above the
specified pullback level, and the next candle to close below the level. We don’t just enter based on the wicks or highs and lows of the candles, they need to close beyond the line for the EA to
consider it to be valid. Once the trigger is met, we would enter immediately at the candle open. I have created this trigger to filter out the noises and spikes in the market so that the signals are
more accurate and fewer will be created. Yes, this would mean that we would miss out on some trades, but I’d rather trade less and take quality setups.
Fibonacci Forex Trading
It’s a semi-automated EA meant to aid traders in the execution aspect of trading. It does all the calculations in finding the optimal lot size, based on the risk, finds its stop loss, and sets the
top as well. You as a trader still have to decide and pick a direction, up or down, and look for areas for entries that can yield positive rewards to risk trades. In the long run, that’s how we stay
in the game as traders, by guessing the right direction more often and winning more on the winning trades. Click here for more info on Chart Pattern Forex.
Discover Fibonacci In nature
The article found at science.howstuffworks.com uncovers the hidden secrets behind the Fibonacci sequence and the golden ratio that radiate the beauty of the universe.
The Fibonacci sequence, first discovered by Indian mathematicians in the 12th century and later documented in a book by an Italian scientist in the 13th century, follows a fascinating and mysterious
pattern of numbers. The numbers in this sequence have a unique relationship where each number is the sum of the two preceding numbers.
Not only that but the Fibonacci sequence is also connected to the golden ratio, which magically appears in various natural phenomena. This ratio, known as the golden ratio or golden mean, has a
consistent value that almost always emerges in awe-inspiring proportions and forms.
Through the Fibonacci sequence and the golden ratio, we can witness the astonishing mathematical harmony that unfolds in the wonders of nature. Here are some examples:
1. Seashells: The spirals in seashells closely resemble the spiral pattern in the Fibonacci sequence.
2. Flower Petals: The number of petals in flowers corresponds to the Fibonacci sequence.
3. Storm Structures: Storm formations, such as tornadoes, exhibit a resemblance to the Fibonacci sequence. The structure of a tornado’s wind can be observed to have a spiral pattern similar to the
Fibonacci sequence.
4. Human Body: Many parts of the human body unknowingly follow the Fibonacci sequence, from the arrangement of facial features to the number of segments in limbs and fingers. The proportions and
sizes of the human body can also be divided using the golden ratio. The DNA molecule follows this sequence, measuring 34 angstroms in length and 21 angstroms in width for each complete double helix
These examples highlight the profound connection between mathematics and nature, demonstrating the remarkable presence of Fibonacci numbers and the golden ratio in various aspects of the natural | {"url":"https://forexjames.com/fibonacci-retracement-forex/","timestamp":"2024-11-03T12:42:56Z","content_type":"text/html","content_length":"59942","record_id":"<urn:uuid:3cad9a0e-43c1-4805-9bd1-bddb79674b84>","cc-path":"CC-MAIN-2024-46/segments/1730477027776.9/warc/CC-MAIN-20241103114942-20241103144942-00843.warc.gz"} |
A three-dimensional forward particle tracking Eulerian Lagrangian localized adjoint method for solution of the contaminant transport equation
The contaminant transport equation is solved in three dimensions using the Eulerian Lagrangian Localized Adjoint Method (ELLAM). Trilinear and finite volume test functions defined by the
characteristics of the governing equation are employed and compared. Integrations are simplified by using forward tracking of integration points and the resultant equations are solved using a
preconditioned conjugate gradient method. The algorithm is coupled to a block-centered finite difference approximation of the groundwater flow equation similar to that used in the popular MODFLOW
code. The ELLAM is tested using a one-dimensional analytical solution and in the simulation of contaminant transport in a three-dimensional variable velocity field. The linear test function ELLAM was
found to be superior to the finite volume ELLAM. Both ELLAM formulations were found to be robust, computationally efficient and relatively straight-forward to implement. When compared to traditional
particle tracking and characteristics codes commonly used in MODFLOW, the ELLAM retains the computational advantages of traditional characteristic methods with the added advantage of good mass
Original language English (US)
Title of host publication Computational methods in water resources - Volume 2 - Computational methods,surface water systems and hydrology
Editors L.R. Bentley, J.F. Sykes, C.A. Brebbia, W.G. Gray, G.F. Pinder, L.R. Bentley, J.F. Sykes, C.A. Brebbia, W.G. Gray, G.F. Pinder
Publisher A.A. Balkema
Pages 611-618
Number of pages 8
ISBN (Print) 9058091252
State Published - 2000
Externally published Yes
Event Computational Methods in Water Resources - Calgary, Canada
Duration: Jun 25 2000 → Jun 29 2000
Other Computational Methods in Water Resources
Country/Territory Canada
City Calgary
Period 6/25/00 → 6/29/00
All Science Journal Classification (ASJC) codes
• General Earth and Planetary Sciences
• General Engineering
• General Environmental Science
Dive into the research topics of 'A three-dimensional forward particle tracking Eulerian Lagrangian localized adjoint method for solution of the contaminant transport equation'. Together they form a
unique fingerprint. | {"url":"https://collaborate.princeton.edu/en/publications/a-three-dimensional-forward-particle-tracking-eulerian-lagrangian","timestamp":"2024-11-13T23:19:11Z","content_type":"text/html","content_length":"52253","record_id":"<urn:uuid:f497c66e-a792-4f0e-871d-12876b486a0c>","cc-path":"CC-MAIN-2024-46/segments/1730477028402.57/warc/CC-MAIN-20241113203454-20241113233454-00807.warc.gz"} |
C.K Raju's "Decolonizing Mathematics"
Decolonizing mathematics | Decolonial International Network (Full Article)
These are just some of the selected quotes
** “In school our children are being taught two conflicting systems of mathematics. In primary school they learn to calculate in an empirical way: one apple and one apple makes two apples. But later
on they they are told that is wrong, and learn some formal mathematics where you postulate some axioms and use the deductive method to arrive at conclusions from the axioms. This makes matters very
complicated: Whitehead and Bertrand Russell took 368 pages to deductively prove 1+1=2 in their book. Decolonised mathematics eliminates this needless complexity and accepts the natural and empirical
way; it is simple and easy.”**
Indeed, that is how I learned mathematics using deductive proofs. The famous example is: “All men are mortal. Socrates is a man. Therefore Socrates is mortal.” So what is wrong with that method of
deduction? Why is this Eurocentric, apart from the fact that a male is used in this example?
Raju: “First of all, I don’t use the term ‘Eurocentric’ because it wrongly suggests that a massive piece of deliberate mischief was an innocent mistake. Second there is nothing wrong with the method
of deduction as such, which was used also, for example, in India. Of course, attributing this syllogism to Aristotle is the usual false Western history: there is nil evidence to link the syllogism to
Aristotle. What is uniquely Western and wrong are the claims that (a) deduction is infallible, (b) that it is universal (c) that deductive proof is superior to empirical proof, and that (d) it is
possible to arrive at valid knowledge without any empirical inputs, as in formal mathematics. All these wrong claims lead to the wrong belief that Western (formal) mathematics is superior and the
only right way to do mathematics.”**
“Empirical proof is rejected by Western mathematics on the grounds that empirical proof is fallible. Our senses might mislead us. To use a classical example from Indian philosophy: I might mistake a
rope for a snake or a snake for a rope. But deductive proof too is fallible: one may easily mistake an invalid deductive proof for a valid one. For example, the very first proposition of “Euclid’s”
Elements has an invalid deductive proof. But for 8 centuries that book was mistakenly regarded by all the foremost minds in the West as the model of deductive proof, when, in fact, there isn’t a
single valid deductive proof in it, as Bertrand Russell too emphasized. How do you know that his own 368 page proof of 1+1=2 is valid? You just blindly trust authority, and such blind trust can be
very fallible. Empirical proofs are never so fallible: one might mistake a rope for a snake, but the Western error about “Euclid’s” Elements, is like mistaking a rope for an elephant.”
“No, on the contrary they are inferior. Divorced from the empirical, even a valid deductive proof does not lead to valid knowledge or even to approximately valid knowledge,” states Raju. “Using the
deductive method any silly proposition whatsoever can be proved as a mathematical theorem from some postulates.”**
to summarize, he is arguing that Western Mathematics reliability on the Deductive method is inferior to the empirical method
Sounds like a joke.
You can’t do “empirical” math. Go ahead, observe the square root of two “in nature.” Ain’t gonna happen.
Now, I do know some people who insist on “constructive math.” They refuse to accept any number that doesn’t represent an actual, physically real number of things. Billions? Sure, there are billions
of people. Trillions? Sure, atoms and molecules and things. But, say, 10^800? “That is not a number!”
I think those guys are jerks, too.
That was straight word salad to me. Is the type of mathematics that he’s railing against the kind that the layman would be educated with ? Or is he talking about a type that is left to math majors
and such ?
The OP is quoting an idiot. An idiot with an axe to grind.
Any thoughts beyond that are casting pearls upon the demand of a swine (the author of the idiot webpage, not the OP)
No Western mathematician that I’m familiar with (at least any who are not kooks) ever made this claim. Indeed, the fact that one can arrive at a false conclusion when starting from invalid axioms is
fundamental to proof theory, and is one of the first things taught in any kind of formal logic course, as well as informal philosophical argumentation courses.
No one claimed that, either.
I think anyone who would make this claim (or wrongly ascribe the claim to a group of people) does not understand what deductive and empirical proof are. One can not be superior to the other, because
they are used for entirely different things. Maybe Raju is arguing that traditional mathematics says that deductive reasoning can accomplish things which empirical observation can not. That’s true.
So is the inverse. They are different tools for different jobs.
What’s “valid knowledge?” Why is knowledge gained from axiomatic, non-empirical systems less valid than other kinds of knowledge?
I think that the way we teach mathematics to kids is badly out of date and curriculums should be redesigned. I don’t think C. K. Raju would be my top choice for doing so.
At the college and graduate school level, there’s plenty of reason why math and science students should study philosophy of mathematics. I was required to take a course in “Math and Society” junior
year, and I can still say it was one of my most important college courses, even though the professor was slightly crazy. Actual mathematicians should know about the deductive vs empirical,
intuitionist vs rationalist controversies, Zorn’s Lemma, the Banach-Tarski paradox, and things like that. But it would be pointless to try pushing those topics into high school, much less elementary
“Decolonizing”. Okaaaay.
Gee, that doesn’t raise any questions about this guy’s issues at all.
Tell me more about this quotation from Raju:
> For example, the very first proposition of “Euclid’s” Elements has an invalid deductive
> proof. But for 8 centuries that book was mistakenly regarded by all the foremost minds
> in the West as the model of deductive proof, when, in fact, there isn’t a single valid
> deductive proof in it, as Bertrand Russell too emphasized.
Let’s see a demonstration that there isn’t a single deductive proof in the Elements.
Silly me, I thought that Indians invented the concept of zero. Silly me, I though that algebra was invented by the Arabs. Silly me.
The evil Western colonists stole it, though. And made it deductive and inferior. :rolleyes:
Yes, correct.
Throughout the history of mathematics, deductive proof and the axiomatic method on the one hand, and empirical verification and real-world application on the other hand, have been complementary.
I wrote:
> a single deductive proof
I meant:
> a single valid deductive proof
I assume this is a reference to the fact that some of Euclid’s proofs (including the first one) rest on assumptions that can’t be justified from his axioms. From the Wikipedia article on Euclidean
The assertion was that there were no valid proofs in Euclid. Not that there was one (or more) invalid proofs.
Wasn’t there a thread some time ago about some academic activist claiming that science and math courses discriminated against women because the answers had to be exact or something?
Same thing here.
I was thinking more of the Alan Sokal article. Good parody should be hard to distinguish from sincerity. I am pretty sure Mr. Raju is serious, but not 100%.
This reads like a homework thread to me, which is against the rules. Thread closed. | {"url":"https://boards.straightdope.com/t/c-k-rajus-decolonizing-mathematics/769884","timestamp":"2024-11-07T09:21:30Z","content_type":"text/html","content_length":"57916","record_id":"<urn:uuid:7886106f-ebd4-4284-976d-f06e85f5b10a>","cc-path":"CC-MAIN-2024-46/segments/1730477027987.79/warc/CC-MAIN-20241107083707-20241107113707-00325.warc.gz"} |
Fastest Way to Generate Random Strings in JavaScript
There are so many ways to generate random strings in JavaScript and it doesn't really matter which method is faster.
The method I like using most is Math.random()
I made a video on it:
Basically the idea is to use Math.random(), then you can convert it to string and do some simple string manipulation on it.
To get random numbers, I would use something like below:
To get random strings with numbers only, I would use:
Math.random().toString().substr(2, 5)
Fortunate .toString() has a param called radix that you can pass in numbers between 2 - 36 which will cast the generated numbers to the radix characters that fall between the given number. The radix
is also known as base and its for representing numeric values
To get a random number between 0-1:
Math.random().toString(2).substr(2, 5)
To get a random number between 0-5:
Math.random().toString(5).substr(2, 5)
Starting from 11/12, it will start introducing letters. So to get a fully random string:
Math.random().toString(20).substr(2, 6)
With this you can now write your awesome random string generator:
const generateRandomString = function(){
return Math.random().toString(20).substr(2, 6)
To be able to change the length of the output:
const generateRandomString = function(length=6){
return Math.random().toString(20).substr(2, length)
One liner
const generateRandomString = (length=6)=>Math.random().toString(20).substr(2, length)
That's all.
If you know of any other faster ways, please I would love to see it in the comment section.
Top comments (7)
wwaterman12 •
Nice function! But one thing I noticed is that this will only work for random alpha-numeric strings up to a certain length. If you want to have something longer, you should call the method
recursively. Something like:
const generateRandomString = function (length, randomString="") {
randomString += Math.random().toString(20).substr(2, length);
if (randomString.length > length) return randomString.slice(0, length);
return generateRandomString(length, randomString);
rohaq •
Yep, looks like it tops out at 12-13 characters for me, demo here: jsfiddle.net/mberwk8a/2/
Phillip Rhodes •
Hi, thanks for the simple, straight-forward article! I just wanted to point out a small error in your one-liner. You define the length variable, but you still used a hard-coded "6" instead of it in
the expression.
diek •
Do you like base20 for something special?
Oyetoke Toby •
No reason actually, I just like using it.
Austin •
Just to clarify for people who don't know, the radix argument for toString goes from 2 to 36, and you need to use 36 to include all alphanumeric characters in the alphabet. Using 20 will omit more
than half of the alphabetic letters from the output.
Jan Mauler •
Don't use the .substr(2, ...) at the end, because you risk having empty string as a result. The result of Math.random() can be 0, so if you substr(2, ...), you will end up with an empty string.
For further actions, you may consider blocking this person and/or reporting abuse | {"url":"https://practicaldev-herokuapp-com.global.ssl.fastly.net/oyetoket/fastest-way-to-generate-random-strings-in-javascript-2k5a","timestamp":"2024-11-08T09:37:35Z","content_type":"text/html","content_length":"144672","record_id":"<urn:uuid:e96b8627-4d6f-4d11-8b99-76b6794f5900>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00532.warc.gz"} |
Using the DCF Method to Value Companies - The British InvestorUsing the DCF Method to Value Companies - The British Investor
Using the DCF Method to Value Companies
Adopting the DCF
I cut my investing teeth on the writings and investment philosophies of value investing legends & authors like Warren Buffett, John Burr Williams, David Dreman and of course, Benjamin Graham. My
favourite investing book is Contrarian Investment Strategies*, first published by Dreman in 1980. Whilst markets, industries and information availability have changed significantly in the decades
since some of these luminaries were active, the fundamental principles of investing with a margin of safety, searching for value, and picking stocks with tangible financial performance have not. It
is this solid grounding that led me to the adoption of the discounted cash flow (DCF) methodology, and I haven’t looked back since.
The Discounted Cash Flow (DCF) Analysis
A discounted cash flow analysis is a tool, nothing more. There are obviously no guarantees of performance/return, the outputs are only as good as the inputs, and in isolation should not be used
solely for investment decisions. There are varying degrees of complexity in establishing the variables that you use on a DCF, however I like to keep mine as simple as possible. The fundamental
premise of the way I do it is thus: I take the cash generating abilities of a business, make my best estimation of the continued ability of that business to grow at a specified rate and duration, and
discount each year’s future cash flows by a rate I determine I could receive with certainty had I invested it elsewhere (which we call the risk-free rate). Because of the increased risk of return in
equity investing, I add an equity risk premium to this risk-free rate to create the discount rate. I add up all the future years’ discounted free cash generation and divide the total by the shares
outstanding to arrive at an intrinsic per-share value.
Simple, but not easy. What I like about it, however, is it helps keep me safe from excessive speculation. What I mean is, by the very virtue of the fact that my method can only be used on businesses
that generate positive cash flow, means I steer well clear of businesses that burn through money with little to show for it, startups who may or may not ever make a profit, or companies that require
such significant investment just to keep their head above water that there remains the distinct possibility that they will eventually fail, or at least run into significant difficulty.
An Example – Associated British Foods (LSE:ABF)
Let’s look at an example using a solid, if not spectacular British FTSE 100 constituent, Associated British Foods PLC.
ABF trades at around 16 times earnings, generates reasonable returns on equity at reasonable margins, and has grown net income at around 5.5% for the past decade. It also generates free cash, year
after year. We firstly need to define our inputs, which generally require three things. One, a determination of free cash flows. For this I take the most recent five year average, which is £802.6m a
year. This helps account for any significant fluctuations, but the most recent year’s free cash flow figure can work too. Next, we look at expected growth rates and duration. I previously mentioned
that net income has grown at 5.5% for a decade, so we’ll use that for the subsequent decade as our growth rate. As standard I use a terminal growth rate of 3% after these ten years, which is a proxy
for the average rate of inflation, and therefore a businesses assumed ability to increase pricing to match.
Finally we need a discount rate. That is, what could we earn with almost certainty in the next ten years (the risk-free rate mentioned earlier) PLUS our equity risk premium. For this I like to use
the 10 year UK Gilt yield, which currently stands at 3.99%. For the equity risk premium I use the long-term average annual return of the respective index, in this case the FTSE 100, which stands at
around 7%. Therefore our discount rate will be 10.99%. We’ll round up to 11%.
The discount rate is the expected, or desired return we want from an investment in the business. Because £100 today is worth more than £100 in one, two or ten years time (due to optionality,
inflation, etc), we therefore discount each future year’s cash flows by our 11% rate to reflect this. The formula looks like this:
If we add up all those future discounted cash flows, dividing by the shares outstanding, we are looking for an intrinsic value above that of the current share price.
So what of Associated British Foods? Putting all those variables into my calculator, I arrive at an intrinsic value of £16.29, compared to the current share price of £22.03 today. This implies that
currently, Associated British Foods is potentially overvalued, and would not represent an enticing place to put my money.
As mentioned before, by going through the DCF process as a starting point (having screened for certain criteria beforehand) I immediately discount the more speculative investments that rely more on
hope than intrinsic analysis. In my view this is a highly underrated step in an assessment process. Secondly, a DCF analysis makes no judgment of perceived “value” based on classic metrics. I.e. a P/
E of 5 is treated the same as a P/E of 50, insofar as a companies ability to grow and generate cash. To take a household name, Amazon has rarely, if ever traded at a low P/E multiple, but has
consistently generated cash, and has created insane returns for long-time holders.
The process also accounts for variances in the wider economic environment without the need for macro-economic forecasting through the variable representing the risk-free rate. In times of ultra low
interest rates, the risk-free rate will be lower, therefore the overall discount rate will be lower, to match. For example, on September 1st 2020, the 10 year Gilt yield was 0.202%. If we used that,
alongside our equity risk premium of 7%, we get a discount rate of 7.202%. Were this the case today, the intrinsic value of ABF would be £29.62, a figure higher than today’s current price and
offering a more compelling argument for investing. Because the risk-free rate is lower, investors look elsewhere for returns, leading to greater stock market performance (in theory), and are willing
to obtain a lower return from equities due to lack of acceptable alternatives. Conversely, when interest rates are high, equities are less appealing and the requirement for adequate return is
The other advantage of the DCF model is that a number of inputs remain fixed. Because of this, it is reasonable, so long as you are realistic in the other variables, to compare two companies
like-for-like to assess where best to put your money. If you are using the same discount rate and the same assumptions on growth beyond your initial period, it makes comparison fairly simple.
The DCF can’t predict the future. You have to make assumptions or estimates, in one way or another, to arrive at any expected or intrinsic value. It does, as the warnings on financial products say,
attempt to predict future returns from past performance. If the notion of this makes you uncomfortable, using a DCF may not be for you.
Because of the variability of return, five people could do a DCF analysis on a company and return five different results, which is why one has to be assured in their risk appetite, expectations, and
understanding of the company under the spotlight. As I said earlier, this is a tool, and should not be taken in isolation without conducting a further assessment of the company, its business model,
and its financial results.
Finally the tool is open to abuse. Were one inclined, one could manipulate the inputs of a favoured company to achieve the desired intrinsic value figure that validates his/her/their opinion or
assumption. It is essential to not use this tool to confirm any pre-existing bias. This is why I often start with a DCF to understand whether I should investigate further, or add a company to a watch
How to Use a DCF
There is of course a significant degree of judgment required to determine those variables you use when calculating a DCF, and that will be open to your own interpretation, or your risk appetite.
Perhaps you want to use the most recent year’s free cash flow figure. Perhaps you want to use a five year average. You may feel your equity risk premium need not be 7%, but that you’d be happy with a
figure of 5%. This would lower your discount rate, and the rising tide of that decision would float all equity boats. All intrinsic equity valuations would be improved because the discount on future
cash flows would be lower.My advice (and it is just advice), would be to investigate further if you feel a tool such as a DCF analysis is something you feel would add to your investing repertoire.
There are models out there that are more complex than my methodology, involving terms like WACC, NPV, etc. But I try not to get in my own way, and instead keep it as simple as my average-IQ brain can
handle. There are also many websites out there that will allow you to use their calculator, or in the very least assist you in building your own DCF model. I heartily recommend Lyn Alden’s page on
Discounted Cash Flow Analysis, which can be found here for more examples.
What I generally look for is those rare companies that appear so mispriced, and return an intrinsic value far higher than their current price, that by design there appears a margin of safety built
in. And thus my original paragraph. If I can find companies with significant perceived discounts to intrinsic value, I can afford to be slightly out in my calculations and still feel reasonably
confident of a good return on my investments over time. My method is far from fully defined and I have years of refinement to go, but as a measure of valuing quality businesses I find it useful to
think of as a guiderail in my search.
Thanks for reading, and happy investing! If you liked this post, please consider sharing to the social media of your choice.
*This is an affiliate link
This Post Has 2 Comments
1. Hi Chriss,
Good summary. I’ve used a similar approach (discounted dividend models) for about five years and I do think these models (cash flow or dividend-based) are useful, especially in terms of coming up
with a stable valuation versus the usual price-earnings ratio approach, which can be affected by the ups and downs of last year’s earnings.
This type of thinking is also used by a couple of my favourite fund managers, Nick Train and also Ian Lance & Nick Purves, both of whom use models that forecast earnings growth out to some future
“normal” year within the medium-term (5-8 years), and then come up with a valuation based on a historically normal multiple of those “normal” future earnings.
It’s different to DCF and DDM models, but in the same ballpark.
Good luck with your return to blogging,
1. Thanks John, appreciate you taking the time to reply and for your kind words.
It’ll be an ongoing process learning how best to construct the DCF, but I think holding on to the principles and purpose of the valuation model is the most important thing. | {"url":"https://thebritishinvestor.com/general-musings/using-the-dcf-method-to-value-companies/","timestamp":"2024-11-03T00:06:50Z","content_type":"text/html","content_length":"85796","record_id":"<urn:uuid:c92ddbfe-1460-4728-98d5-db9c437899de>","cc-path":"CC-MAIN-2024-46/segments/1730477027768.43/warc/CC-MAIN-20241102231001-20241103021001-00207.warc.gz"} |
Exterior Angle Theorem And Triangle Sum Theorem Worksheet - Angleworksheets.com
Exterior Angle Theorem And Triangle Sum Theorem Worksheet – This article will discuss Angle Triangle Worksheets as well as the Angle Bisector Theorem. In addition, we’ll talk about Isosceles and
Equilateral triangles. You can use the search bar to locate the worksheet you are looking for if you aren’t sure. Angle Triangle Worksheet This Angle … Read more | {"url":"https://www.angleworksheets.com/tag/exterior-angle-theorem-and-triangle-sum-theorem-worksheet/","timestamp":"2024-11-08T20:55:47Z","content_type":"text/html","content_length":"47626","record_id":"<urn:uuid:77e11332-d096-4282-a58f-89f91ba295e7>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00621.warc.gz"} |
Everybody talks about Monads when they mention Haskell, so I got a bit ahead of myself and wanted to see something of what they’re about. No, don’t worry, I’m not aspiring to yet another Monad
tutorial. I feel I have a ways to go before I’m ready to craft my own light-saber.
I did read about 10 Monad articles on the Web, and found myself more confused when I came out than when I went in. Today’s exercise took about 5-6 hours of pure frustration, before a kind soul on IRC
finally set me straight. It sure is difficult when getting past a single compiler error takes you hours.
That bedeviled cat
Most geeks know about Schrödinger’s cat, the fated beast who, when put into a box with a random source tied to a deadly gas trigger, remains in a state of quantum superposition in which he’s neither
alive nor dead until someone opens the box to look.
Well, people kept saying that Monad are like “computational containers”, so I wanted to model the following:
1. There is a Schroedinger Monad into which you can put a Cat.
2. When you create the Monad, it is Unopened, and the Cat’s has no state.
3. You also pass in a random generator from the outside world. This involves another Monad, the IO Monad, because randomness relates to the “world outside”.
4. As long as you don’t use the monad object, the Cat’s is neither Dead nor Live.
5. As soon as you peek into the box, or use it in any calculation, the Cat’s fate is decided by a roll of the dice.
When I run the program ten times in a row, here’s what I get:
Opened (Live (Cat "Felix"))
Opened Dead
Opened Dead
Opened Dead
Opened Dead
Opened (Live (Cat "Felix"))
Opened Dead
Opened Dead
Opened (Live (Cat "Felix"))
Opened (Live (Cat "Felix"))
Let’s look at the code, and where I had troubles writing it.
A flip of the coin
The first function flips a coin and returns True or False to represent Heads or Tails:
The sugar fst $ random gen is just shorthand for fst (random gen). There is no difference, I was just playing with syntax. You do need to pass in a valid random generator, of type StdGen, for the
function to work.
These two types let me make Cats out of Strings, along with a Probable type which models a Live thing or a Dead thing. It treats all Dead things as equal. I can create a Live Cat with:
Following my “fun with syntax” up above, I could also have written:
It doesn’t matter which. The $ character is the same as space, but with much lower precedence so that parentheses aren’t needed around the argument. If there were no parens, it would look like I was
calling Live with two separate arguments: Cat and "Felix".
Flipping a Cat
When I have a Cat, I can subject it to a coin toss in order to get back a Live Cat or a Dead one. I should probably have called this function randomGasTrigger, but hey.
The type of the function says that it expects a random generator (for flipCoin), some thing, and returns a Probable instance of that thing. The Probable means “can be Live or Dead”, according to how
I defined the type above. The rest of the function is pretty clear, since it looks a lot like its imperative cousin would have.
Bringing in Schroedinger
This type declaration is more complicated. It creates a Schroedinger type which has two data constructors: an Opened constructor which takes a Probable object – that is, whose Live or Dead state is
known – and an Unopened constructor which takes a random generator, and an object without a particular state, such as a Cat.
Some values I could create with this type:
felix = Opened (Live (Cat "Felix")) -- lucky Felix
poorGuy = Opened Dead -- DOA
unknown = Unopened (mkStdGen 100) (Cat "Felix")
In the third case, the idea is that his fate will be determined by the random generator created with mkStdGen 100. However, I want a real random source, so I’m going to get one from the environment
Here comes the Monad
instance Monad Schroedinger where
Opened Dead >>= _ = Opened Dead
Opened (Live a) >>= f = f a
Unopened y x >>= f = Opened (flipCat y x) >>= f
return x = Opened (Live x)
As complex as Monads sound on the Web, they are trivial to define. Maybe it’s a lot like binary code: nothing could be simpler than ones and zeroes, yet consider that all complexity expressable by
computers, down to video, audio, programming languages, and reading this article, are contained within the possibilities of those two digits. Yeah. Monads are a little like that.
This useless Monad just illustrates how to define one, so let’s cut it apart piece by piece. By the way, I didn’t author this thing, I just started it. Much of its definition was completed by folks
on IRC, who had to wipe the drool from my face toward the end.
Says that my Schroedinger type now participates in the joy and fun of Monads! He can be discussed at parties with much auspiciousness.
The >>= operator is the “bind” function. It happens when you bind a function to a Monad, which is like applying a function to it. This line says that if you apply a function to an Opened box
containing a Dead thing, what you’ll get back is an Opened box with a Dead thing.
If, however, you bind a function to an Opened box with a Live thing, it will apply the function to what’s in the box – in this case, the Cat itself. The function f is assumed to return another
instance of the Schroedinger type, most likely containing the same cat or some transformed version of it.
Here is the meat of this example, it’s reason for being, all contained within this one line: If you bind a function to an Unopened box, it gets bound in turn to an Opened box containing a Cat whose
fate has been decided by the dice. That’s all. The reason I used a Monad to do this is to defer the cat’s fate until someone actually looked inside the container.
Lastly, if someone returns a cat from a box, assume its an Opened box with a Live Cat. I don’t honestly understand why this is necessary, but it seems Opened Dead cats are handled by the binding
above, as shown by the output from my program. I’ll have to figure this part out soon…
The main function
The last part of the example is the main routine:
main = do
gen ,- getStdGen
print (do
box ,- Unopened gen (Cat "Felix")
-- The cat's fate is undecided
return box)
This is fairly linear: it gets a random generator from the operating system, then creates an Unopened box and returns it, which gets printed. print does its work by calling show on the Schroedinger
type, since it was derived from Show earlier.
Something I still don’t understand: at exactly which point does the flipping happen? When box is returned? When show gets called? Or when print actually needs the value from show in order to pass it
out to the IO subsystem?
Closing thoughts
The full version of this code is on my server. There is also a simpler version without Monads. I worked on the Monad version just to tweak my brain. At least I can say I’m closer to understanding
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Tal Moran
I am a faculty member in the School of Computer Science at Reichman University (formerly IDC Herzliya).
Before joining Reichman University, I was a postdoctoral research fellow at the Center for Research on Computation and Society at Harvard University advised by Salil Vadhan.
I did my Phd at The Weizmann Institute of Science under the supervision of Moni Naor.
My research focuses on employing ideas and techniques from theoretical cryptography to design secure systems in the "real world". I'm also interested the foundations of cryptography and the theory of
cryptography in general.
A great example of where theoretical cryptography can help in the real world is in the design of crypto-currencies, such as Bitcoin, Ethereum and Spacemesh (whose design is based on my own research).
At the heart of every (good) crypto-currency lies a cryptographic protocol for distributed "permissionless" consensus. Such protocols are already being used in the real world, and pose fascinating
theoretical questions—with connections to distributed systems, multi-party computation and more.
Another interesting example is the development of cryptographic protocols for "end-to-end verifiable" (E2E), secure elections. In an E2E election, voters can be certain that their votes were
correctly tabulated even if they don't trust the computers used to run the election! At the same time, a voter cannot prove to anyone else how they voted, making coercion and vote-buying more
Eurocrypt 2019
: May 19–23, 2019, Darmstadt
ACM CCS 2018
: Oct. 15–19, 2018, Toronto
ACM CCS 2017
: Oct. 30–Nov. 3, 2017, Dallas
CT-RSA 2016
: Feb. 29–Mar. 4, 2016, San Francisco
VoteID 2015
: Sep 2–4, 2015, Bern
CT-RSA 2014
: Feb. 24–28, 2014, San Francisco
Crypto 2013
: Aug. 18–22, 2013, Santa Barbara
EVT/WOTE 2012
: Aug. 6–7, 2012, Bellevue
VoteID 2011
: Sep. 28–30, 2011, Tallinn
EVT/WOTE 2011
: Aug. 8–9, 2011, San Francisco
Crypto 2011
: Aug. 14–18, 2013, Santa Barbara
CT-RSA 2011
: Feb. 14–18, 2011, San Francisco
Crypto 2010
: Aug. 15–19, 2013, Santa Barbara
EVT/WOTE 2009
(PC co-chair): Aug. 10–11, 2009, Montreal
WOTE '07
: Jun. 20–21, 2007, Ottawa
• Iddo Bentov, Pavel Hubácek, Tal Moran and Asaf Nadler
Tortoise and Hares Consensus: The Meshcash Framework for Incentive-compatible, Scalable Cryptocurrencies
CSCML 2021 [BibTex]
author = {Iddo Bentov and Pavel Hub\'{a}cek and Tal Moran and Asaf Nadler},
title = {Tortoise and Hares Consensus: The Meshcash Framework for Incentive-compatible, Scalable Cryptocurrencies},
editor = {Shlomi Dolev and Oded Margalit and Benny Pinkas and Alexander Schwarzmann},
booktitle = {CSCML 2021},
pages = {114--127},
series = {Lecture Notes in Computer Science},
volume = {12716},
year = {2021},
publisher = {Springer},
url = {https://eprint.iacr.org/2017/300.pdf},
In this paper, we propose Meshcash, a protocol for implementing a permissionless ledger (blockchain). Unlike most existing proof-of-work based consensus protocols, Meshcash does not rely on
leader-election (e.g., the single miner who managed to extend the longest chain). Rather, we use ideas from traditional (permissioned) Byzantine agreement protocols in a novel way to guarantee
convergence to a consensus from any starting state. Our construction combines a local “hare” protocol that guarantees fast consensus on recent blocks (but doesn't, by itself, imply
irreversibility) with a global “tortoise” protocol that guarantees irreversibility. Our global protocol also allows the ledger to “self-heal” from arbitrary violations of the security
assumptions, reconverging to consensus after the assumptions hold again. Meshcash is designed to be race-free: there is no “race” to generate the next block and honestly-generated blocks are
always rewarded. This property, which we define formally as a game-theoretic notion, turns out to be useful in analyzing rational miners' behavior: we prove (using a generalization of the
blockchain mining games of Kiayias et al.) that race-free blockchain protocols are incentive-compatible and satisfy linearity of rewards (i.e., a party receives rewards proportional to its
computational power).
Because Meshcash can tolerate a high block rate regardless of network propagation delays (which will only affect latency), it allows us to lower both the variance and the expected time between
blocks for honest miners; together with linearity of rewards, this makes pooled mining far less attractive. Moreover, race-free protocols scale more easily (in terms of transaction rate). This is
because the race-free property implies that the network propagation delays are not a factor in terms of rewards, which removes the main impediment to accommodating a larger volume of
transactions. We formally prove that all of our guarantees hold in the bounded-delay communication model of Pass, Seeman and shelat, and against a constant fraction of Byzantine (malicious)
miners; not just rational ones.
• Chen-Da Liu-Zhang, Julian Loss, Ueli Maurer, Tal Moran and Daniel Tschudi
MPC with Synchronous Security and Asynchronous Responsiveness
Asiacrypt 2020 [BibTex]
author = {Chen{-}Da Liu{-}Zhang and Julian Loss and Ueli Maurer and Tal Moran and Daniel Tschudi},
title = {MPC with Synchronous Security and Asynchronous Responsiveness},
editor = {Shiho Moriai and Huaxiong Wang},
booktitle = {Asiacrypt 2020},
pages = {92--119},
series = {Lecture Notes in Computer Science},
volume = {12493},
year = {2020},
month = {December},
publisher = {Springer},
url = {https://eprint.iacr.org/2019/159.pdf},
Two paradigms for secure MPC are synchronous and asynchronous protocols. While synchronous protocols tolerate more corruptions and allow every party to give its input, they are very slow because
the speed depends on the conservatively assumed worst-case delay $\Delta$ of the network. In contrast, asynchronous protocols allow parties to obtain output as fast as the actual network allows,
a property called responsiveness, but unavoidably have lower resilience and parties with slow network connections cannot give input.
It is natural to wonder whether it is possible to leverage synchronous MPC protocols to achieve responsiveness, hence obtaining the advantages of both paradigms: full security with responsiveness
up to $t$ corruptions, and extended security (full security or security with unanimous abort) with no responsiveness up to $T \ge t$ corruptions. We settle the question by providing matching
feasibility and impossibility results:
□ For the case of unanimous abort as extended security, there is an MPC protocol if and only if $T + 2t < n$.
□ For the case of full security as extended security, there is an MPC protocol if and only if $T < \frac{n}{2}$ and $T + 2t < n$. In particular, setting $t = \frac{n}{4}$ allows to achieve a
fully secure MPC for honest majority, which in addition benefits from having substantial responsiveness.
• Marshall Ball, Elette Boyle, Ran Cohen, Lisa Kohl, Tal Malkin, Pierre Meyer and Tal Moran
Topology-Hiding Communication from Minimal Assumptions
TCC 2020 [BibTex]
author = {Marshall Ball and Elette Boyle and Ran Cohen and Lisa Kohl and Tal Malkin and Pierre Meyer and Tal Moran},
title = {Topology-Hiding Communication from Minimal Assumptions},
editor = {Rafael Pass and Krzysztof Pietrzak},
booktitle = {TCC 2020},
pages = {473--501},
series = {Lecture Notes in Computer Science},
volume = {12551},
year = {2020},
month = {November},
publisher = {Springer},
url = {https://eprint.iacr.org/2021/388},
Topology-hiding broadcast (THB) enables parties communicating over an incomplete network to broadcast messages while hiding the topology from within a given class of graphs. THB is a central tool
underlying general topology-hiding secure computation (THC) (Moran et al. TCC'15). Although broadcast is a privacy-free task, it was recently shown that THB for certain graph classes necessitates
computational assumptions, even in the semi-honest setting, and even given a single corrupted party.
In this work we investigate the minimal assumptions required for topology-hiding communication—both Broadcast or Anonymous Broadcast (where the broadcaster's identity is hidden). We develop new
techniques that yield a variety of necessary and sufficient conditions for the feasibility of THB/THAB in different cryptographic settings: information theoretic, given existence of key
agreement, and given existence of oblivious transfer. Our results show that feasibility can depend on various properties of the graph class, such as connectivity, and highlight the role of
different properties of topology when kept hidden, including direction, distance, and/or distance-of-neighbors to the broadcaster.
An interesting corollary of our results is a dichotomy for THC with a public number of at least three parties, secure against one corruption: information-theoretic feasibility if all graphs are
2-connected; necessity and sufficiency of key agreement otherwise.
• Tal Moran and Daniel Wichs
Incompressible Encodings
Crypto 2020 [BibTex]
author = {Tal Moran and Daniel Wichs},
title = {Incompressible Encodings},
booktitle = {Crypto 2020},
pages = {494--523},
series = {Lecture Notes in Computer Science},
volume = {12171},
year = {2020},
month = {August},
url = {https://eprint.iacr.org/2020/814.pdf},
An incompressible encoding can probabilistically encode some data $m$ into a codeword $c$, which is not much larger. Anyone can decode the codeword $c$ to recover the original data $m$. However,
the codeword $c$ cannot be efficiently compressed, even if the original data $m$ is given to the decompression procedure on the side. In other words, $c$ is an efficiently decodable
representation of $m$, yet is computationally incompressible even given $m$. An incompressible encoding is composable if many encodings cannot be simultaneously compressed. The recent work of
Damgård, Ganesh and Orlandi (CRYPTO '19) defined a variant of incompressible encodings as a building block for “proofs of replicated storage”. They constructed incompressible encodings in an
ideal permutation model, but it was left open if they can be constructed under standard assumptions, or even in the more basic random-oracle model. In this work, we undertake the comprehensive
study of incompressible encodings as a primitive of independent interest and give new constructions, negative results and applications:
□ We construct incompressible encodings in the common random string (CRS) model under either Decisional Composite Residuosity (DCR) or Learning with Errors (LWE). However, the construction has
several drawbacks: (1) it is not composable, (2) it only achieves selective security, and (3) the CRS is as long as the data $m$.
□ We leverage the above construction to also get a scheme in the random-oracle model, under the same assumptions, that avoids all of the above drawbacks. Furthermore, it is significantly more
efficient than the prior ideal-model construction.
□ We give black-box separations, showing that incompressible encodings in the plain model cannot be proven secure under any standard hardness assumption, and incompressible encodings in the CRS
model must inherently suffer from all of the drawbacks above.
□ We give a new application to “big-key cryptography in the bounded-retrieval model”, where secret keys are made intentionally huge to make them hard to exfiltrate. Using incompressible
encodings, we can get all the security benefits of a big key without wasting storage space, by having the key to encode useful data.
• Rio LaVigne, Chen-Da Liu-Zhang, Ueli Maurer, Tal Moran, Marta Mularczyk and Daniel Tschudi
Topology-Hiding Computation for Networks with Unknown Delays
PKC 2020 [BibTex]
author = {Rio LaVigne and Chen-Da Liu-Zhang and Ueli Maurer and Tal Moran and Marta Mularczyk and Daniel Tschudi},
title = {Topology-Hiding Computation for Networks with Unknown Delays},
booktitle = {PKC 2020},
pages = {215--245},
series = {Lecture Notes in Computer Science},
volume = {12111},
year = {2020},
month = {June},
publisher = {Springer},
url = {https://eprint.iacr.org/2019/1211.pdf},
Topology-Hiding Computation (THC) allows a set of parties to securely compute a function over an incomplete network without revealing information on the network topology. Since its introduction
in TCC'15 by Moran et al., the research on THC has focused on reducing the communication complexity, allowing larger graph classes, and tolerating stronger corruption types.
All of these results consider a fully synchronous model with a known upper bound on the maximal delay of all communication channels. Unfortunately, in any realistic setting this bound has to be
extremely large, which makes all fully synchronous protocols inefficient. In the literature on multi-party computation, this is solved by considering the fully asynchronous model. However, THC is
unachievable in this model (and even hard to define), leaving even the definition of a meaningful model as an open problem.
The contributions of this paper are threefold.
First, we introduce a meaningful model of unknown and random communication delays for which THC is both definable and achievable. The probability distributions of the delays can be arbitrary for
each channel, but one needs to make the (necessary) assumption that the delays are independent. The existing fully-synchronous THC protocols do not work in this setting and would, in particular,
leak information about the topology.
Second, in the model with trusted stateless hardware boxes introduced at Eurocrypt'18 by Ball et al., we present a THC protocol that works for any graph class.
Third, we explore what is achievable in the standard model without trusted hardware and present a THC protocol for specific graph types (cycles and trees) secure under the DDH assumption.
The speed of all protocols scales with the actual (unknown) delay times, in contrast to all previously known THC protocols whose speed is determined by the assumed upper bound on the network
• Adi Akavia, Rio LaVigne and Tal Moran
Topology-Hiding Computation on All Graphs
J. Cryptology, 33(1):176–227, 2020 [BibTex]
author = {Adi Akavia and Rio LaVigne and Tal Moran},
title = {Topology-Hiding Computation on All Graphs},
pages = {176--227},
volume = {33},
number = {1},
year = {2020},
month = {January},
url = {https://eprint.iacr.org/2017/296.pdf},
journal = {J. Cryptology},
doi = {10.1007/s00145-019-09318-y},
A distributed computation in which nodes are connected by a partial communication graph is called topology-hiding if it does not reveal information about the graph beyond what is revealed by the
output of the function. Previous results have shown that topology-hiding computation protocols exist for graphs of constant degree and logarithmic diameter in the number of nodes
[Moran-Orlov-Richelson, TCC'15; Hirt et al., Crypto'16] as well as for other graph families, such as cycles, trees, and low circumference graphs [Akavia-Moran, Eurocrypt'17], but the feasibility
question for general graphs was open.
In this work we positively resolve the above open problem: we prove that topology-hiding computation is feasible for all graphs under either the Decisional Diffie-Hellman or Quadratic-Residuosity
Our techniques employ random or deterministic walks to generate paths covering the graph, upon which we apply the Akavia-Moran topology-hiding broadcast for chain-graphs (paths). To prevent
topology information revealed by the random-walk, we design multiple graph-covering sequences that, together, are locally identical to receiving at each round a message from each neighbor and
sending back a processed message from some neighbor (in a randomly permuted order).
[Preliminary version appeared in Crypto 2017 |
author = {Adi Akavia and Rio LaVigne and Tal Moran},
title = {Topology-Hiding Computation for All Graphs},
booktitle = {Crypto 2017},
pages = {447--467},
series = {Lecture Notes in Computer Science},
volume = {10401},
year = {2017},
month = {August},
publisher = {Springer},
url = {https://eprint.iacr.org/2017/296},
doi = {10.1007/978-3-319-63688-7_15},
• Tal Moran and Ilan Orlov
Simple Proofs of Spacetime and Rational Proofs of Storage
Crypto 2019 [BibTex]
author = {Tal Moran and Ilan Orlov},
title = {Simple Proofs of Spacetime and Rational Proofs of Storage},
booktitle = {Crypto 2019},
pages = {381--409},
series = {Lecture Notes in Computer Science},
volume = {11692},
year = {2019},
publisher = {Springer},
url = {https://eprint.iacr.org/2016/035.pdf},
doi = {10.1007/978-3-030-26948-7\_14},
We introduce a new cryptographic primitive: Proofs of Space-Time (PoSTs) and construct an extremely simple, practical protocol for implementing these proofs. A PoST allows a prover to convince a
verifier that she spent a “space-time” resource (storing data—space—over a period of time). Formally, we define the PoST resource as a trade-off between CPU work and space-time (under reasonable
cost assumptions, a rational user will prefer to use the lower-cost space-time resource over CPU work).
Compared to a proof-of-work, a PoST requires less energy use, as the “difficulty” can be increased by extending the time period over which data is stored without increasing computation costs. Our
definition is very similar to “Proofs of Space” [ePrint 2013/796, 2013/805] but, unlike the previous definitions, takes into account amortization attacks and storage duration. Moreover, our
protocol uses a very different (and much simpler) technique, making use of the fact that we explicitly allow a space-time tradeoff, and doesn't require any non-standard assumptions (beyond random
oracles). Unlike previous constructions, our protocol allows incremental difficulty adjustment, which can gracefully handle increases in the price of storage compared to CPU work. In addition, we
show how, in a crypto-currency context, the parameters of the scheme can be adjusted using a market-based mechanism, similar in spirit to the difficulty adjustment for PoW protocols.
• Marshall Ball, Elette Boyle, Ran Cohen, Tal Malkin and Tal Moran
Is Information-Theoretic Topology-Hiding Computation Possible?
TCC 2019 [BibTex]
author = {Marshall Ball and Elette Boyle and Ran Cohen and Tal Malkin and Tal Moran},
title = {Is Information-Theoretic Topology-Hiding Computation Possible?},
editor = {Dennis Hofheinz and Alon Rosen},
booktitle = {TCC 2019},
pages = {502--530},
series = {Lecture Notes in Computer Science},
volume = {11891},
year = {2019},
month = {November},
publisher = {Springer},
url = {https://eprint.iacr.org/2019/1094.pdf},
doi = {10.1007/978-3-030-36030-6\_20},
Topology-hiding computation (THC) is a form of multi-party computation over an incomplete communication graph that maintains the privacy of the underlying graph topology. Existing THC protocols
consider an adversary that may corrupt an arbitrary number of parties, and rely on cryptographic assumptions such as DDH.
In this paper we address the question of whether information-theoretic THC can be achieved by taking advantage of an honest majority. In contrast to the standard MPC setting, this problem has
remained open in the topology-hiding realm, even for simple “privacy-free” functions like broadcast, and even when considering only semi-honest corruptions.
We uncover a rich landscape of both positive and negative answers to the above question, showing that what types of graphs are used and how they are selected is an important factor in determining
the feasibility of hiding topology information-theoretically. In particular, our results include the following.
□ We show that topology-hiding broadcast (THB) on a line with four nodes, secure against a single semi-honest corruption, implies key agreement. This result extends to broader classes of
graphs, e.g., THB on a cycle with two semi-honest corruptions.
□ On the other hand, we provide the first feasibility result for information-theoretic THC: for the class of cycle graphs, with a single semi-honest corruption.
Given the strong impossibilities, we put forth a weaker definition of distributional-THC, where the graph is selected from some distribution (as opposed to worst-case).
□ We present a formal separation between the definitions, by showing a distribution for which information theoretic distributional-THC is possible, but even topology-hiding broadcast is not
possible information-theoretically with the standard definition.
□ We demonstrate the power of our new definition via a new connection to adaptively secure low-locality MPC, where distributional-THC enables parties to “reuse” a secret low-degree
communication graph even in the face of adaptive corruptions.
• Rio LaVigne, Chen-Da Liu-Zhang, Ueli Maurer, Tal Moran, Marta Mularczyk and Daniel Tschudi
Topology-Hiding Computation Beyond Semi-Honest Adversaries
TCC 2018 [BibTex]
author = {Rio LaVigne and Chen{-}Da Liu{-}Zhang and Ueli Maurer and Tal Moran and Marta Mularczyk and Daniel Tschudi},
title = {Topology-Hiding Computation Beyond Semi-Honest Adversaries},
booktitle = {TCC 2018},
pages = {3--35},
series = {Lecture Notes in Computer Science},
volume = {11240},
year = {2018},
month = {November},
publisher = {Springer},
url = {https://eprint.iacr.org/2018/255.pdf},
doi = {10.1007/978-3-030-03810-6\_1},
Topology-hiding communication protocols allow a set of parties, connected by an incomplete network with unknown communication graph, where each party only knows its neighbors,to construct a
complete communication network such that the network topology remains hidden even from a powerful adversary who can corrupt parties. This communication network can then be used to perform
arbitrary tasks, for example secure multi-party computation, in a topology-hiding manner.
Previously proposed protocols could only tolerate passive corruption. This paper proposes protocols that can also tolerate fail-corruption (i.e., the adversary can crash any party at any point in
time) and so-called semi-malicious corruption (i.e., the adversary can control a corrupted party’s randomness), without leaking more than an arbitrarily small fraction of a bit of information
about the topology. A small-leakage protocol was recently proposed by Ball et al. [Eurocrypt’18], but only under the unrealistic set-up assumption that each party has a trusted hardware module
containing secret correlated pre-set keys, and with the further two restrictions that only passively corrupted parties can be crashed by the adversary, and semi-malicious corruption is not
tolerated. Since leaking a small amount of information is unavoidable, as is the need to abort the protocol in case of failures, our protocols seem to achieve the best possible goal in a model
with fail-corruption. Further contributions of the paper are applications of the protocol to obtain secure MPC protocols, which requires a way to bound the aggregated leakage when multiple
small-leakage protocols are executed in parallel or sequentially. Moreover, while previous protocols are based on the DDH assumption, a new so-called PKCR public-key encryption scheme based on
the LWE assumption is proposed, allowing to base topology-hiding computation on LWE. Furthermore, a protocol usingfully-homomorphic encryption achieving very low round complexity is proposed.
• Marshall Ball, Elette Boyle, Tal Malkin and Tal Moran
Exploring the Boundaries of Topology-Hiding Computation
Eurocrypt 2018 [BibTex]
author = {Marshall Ball and Elette Boyle and Tal Malkin and Tal Moran},
title = {Exploring the Boundaries of Topology-Hiding Computation},
booktitle = {Eurocrypt 2018},
pages = {294--325},
series = {Lecture Notes in Computer Science},
volume = {10822},
year = {2018},
month = {April},
publisher = {Springer},
doi = {10.1007/978-3-319-78372-7_10},
Topology-hiding computation (THC) is a form of multi-party computation over an incomplete communication graph that maintains the privacy of the underlying graph topology. In a line of recent
works [Moran, Orlov & Richelson, TCC'15, Hirt et al. CRYPTO'16, Akavia & Moran EUROCRYPT'17, Akavia et al. CRYPTO'17], THC protocols for securely computing any function in the semi-honest setting
have been constructed. In addition, it was shown by Moran et al. that in the fail-stop setting THC with negligible leakage on the topology is impossible.
In this paper, we further explore the feasibility boundaries of THC.
□ We show that even against semi-honest adversaries, topology-hiding broadcast on a small (4-node) graph implies oblivious transfer; in contrast, trivial broadcast protocols exist
unconditionally if topology can be revealed.
□ We strengthen the lower bound of Moran et al., identifying and extending a relation between the amount of leakage on the underlying graph topology that must be revealed in the fail-stop
setting, as a function of the number of parties and communication round complexity: Any $n$-party protocol leaking $\delta$ bits for $\delta \in (0,1]$ must have $\Omega(n/\delta)$ rounds.
We then present THC protocols providing close-to-optimal leakage rates, for unrestricted graphs on $n$ nodes against a fail-stop adversary controlling a dishonest majority of the $n$ players.
These constitute the first general fail-stop THC protocols. Specifically, for this setting we show:
□ A THC protocol that leaks at most one bit and requires $O(n^2)$ rounds.
□ A THC protocol that leaks at most $\delta$ bits for arbitrarily small non-negligible $\delta$, and requires $O(n^3/\delta)$ rounds.
These protocols also achieve full security (with no leakage) for the semi-honest setting. Our protocols are based on one-way functions and a (stateless) secure hardware box primitive. This
provides a theoretical feasibility result, a heuristic solution in the plain model using general-purpose obfuscation candidates, and a potentially practical approach to THC via commodity hardware
such as Intel SGX. Interestingly, even with such hardware, proving security requires sophisticated simulation techniques.
• Adi Akavia and Tal Moran
Topology-Hiding Computation Beyond Logarithmic Diameter
Eurocrypt 2017 [BibTex]
author = {Adi Akavia and Tal Moran},
title = {Topology-Hiding Computation Beyond Logarithmic Diameter},
booktitle = {Eurocrypt 2017},
pages = {609--637},
series = {Lecture Notes in Computer Science},
volume = {10212},
year = {2017},
month = {April},
url = {https://eprint.iacr.org/2017/130.pdf},
doi = {10.1007/978-3-319-56617-7},
A distributed computation in which nodes are connected by a partial communication graph is called topology-hiding if it does not reveal information about the graph (beyond what is revealed by the
output of the function). Previous results [Moran, Orlov, Richelson; TCC'15] have shown that topology-hiding computation protocols exist for graphs of logarithmic diameter (in the number of
nodes), but the feasibility question for graphs of larger diameter was open even for very simple graphs such as chains, cycles and trees.
In this work, we take a step towards topology-hiding computation protocols for arbitrary graphs by constructing protocols that can be used in a large class of large-diameter networks, including
cycles, trees and graphs with logarithmic circumference. Our results use very different methods from [MOR15] and can be based on a standard assumption (such as DDH).
• Tal Moran, Moni Naor and Gil Segev
An Optimally Fair Coin Toss
J. Cryptology, 29(3):491–513, 2016 [BibTex]
author = {Tal Moran and Moni Naor and Gil Segev},
title = {An Optimally Fair Coin Toss},
pages = {491--513},
volume = {29},
number = {3},
year = {2016},
url = {http://dx.doi.org/10.1007/s00145-015-9199-z},
journal = {J. Cryptology},
doi = {10.1007/s00145-015-9199-z},
We address one of the foundational problems in cryptography: the bias of coin-flipping protocols. Coin-flipping protocols allow mutually distrustful parties to generate a common unbiased random
bit, guaranteeing that even if one of the parties is malicious, it cannot significantly bias the output of the honest party. A classical result by Cleve (Proceedings of the 18th annual ACM
symposium on theory of computing, pp 364–369, 1986) showed that for any two-party $r$-round coin-flipping protocol there exists an efficient adversary that can bias the output of the honest party
by $\Omega(1/r)$. However, the best previously known protocol only guarantees $O(1/\sqrt{r})$ bias, and the question of whether Cleve?s bound is tight has remained open for more than 20 years. In
this paper, we establish the optimal trade-off between the round complexity and the bias of two-party coin-flipping protocols. Under standard assumptions (the existence of oblivious transfer), we
show that Cleve's lower bound is tight: We construct an $r$-round protocol with bias $O(1/r)$.
[Preliminary version appeared in TCC 2009 |
author = {Tal Moran and Moni Naor and Gil Segev},
title = {An Optimally Fair Coin Toss},
booktitle = {TCC 2009},
pages = {1--18},
series = {Lecture Notes in Computer Science},
volume = {5444},
year = {2009},
month = {March},
publisher = {Springer},
• Yiling Chen, Stephen Chong, Ian A. Kash, Tal Moran and Salil P. Vadhan
Truthful Mechanisms for Agents That Value Privacy
ACM Trans. Economics and Comput., 4(3):13:1–13:30, 2016 [BibTex]
author = {Yiling Chen and Stephen Chong and Ian A. Kash and Tal Moran and Salil P. Vadhan},
title = {Truthful Mechanisms for Agents That Value Privacy},
pages = {13:1--13:30},
volume = {4},
number = {3},
year = {2016},
url = {http://doi.acm.org/10.1145/2892555},
journal = {ACM Trans. Economics and Comput.},
doi = {10.1145/2892555},
Recent work has constructed economic mechanisms that are both truthful and differentially private. In these mechanisms, privacy is treated separately from truthfulness; it is not incorporated in
players' utility functions (and doing so has been shown to lead to nontruthfulness in some cases). In this work, we propose a new, general way of modeling privacy in players' utility functions.
Specifically, we only assume that if an outcome $o$ has the property that any report of player $i$ would have led to $o$ with approximately the same probability, then $o$ has a small privacy cost
to player $i$. We give three mechanisms that are truthful with respect to our modeling of privacy: for an election between two candidates, for a discrete version of the facility location problem,
and for a general social choice problem with discrete utilities (via a VCG-like mechanism). As the number $n$ of players increases, the social welfare achieved by our mechanisms approaches
optimal (as a fraction of $n$).
[Preliminary version appeared in EC 2013 |
author = {Yiling Chen and Stephen Chong and Ian A. Kash and Tal Moran and Salil Vadhan},
title = {Truthful Mechanisms for Agents that Value Privacy},
editor = {Michael Kearns and R. Preston McAfee and {\'E}va Tardos},
booktitle = {EC 2013},
pages = {215--232},
year = {2013},
month = {June},
publisher = {ACM},
• Tal Moran, Ilan Orlov and Silas Richelson
Topology-Hiding Computation
TCC 2015 [BibTex]
author = {Tal Moran and Ilan Orlov and Silas Richelson},
title = {Topology-Hiding Computation},
editor = {Yevgeniy Dodis and Jesper Buus Nielsen},
booktitle = {TCC 2015},
pages = {169--198},
series = {Lecture Notes in Computer Science},
volume = {9014},
year = {2015},
publisher = {Springer},
ee = {\url{https://eprint.iacr.org/2014/1022}},
Secure Multi-party Computation (MPC) is one of the foundational achievements of modern cryptography, allowing multiple, distrusting, parties to jointly compute a function of their inputs, while
revealing nothing but the output of the function. Following the seminal works of Yao and Goldreich, Micali and Wigderson and Ben-Or, Goldwasser and Wigderson, the study of MPC has expanded to
consider a wide variety of questions, including variants in the attack model, underlying assumptions, complexity and composability of the resulting protocols.
One question that appears to have received very little attention, however, is that of MPC over an underlying communication network whose structure is, in itself, sensitive information. This
question, in addition to being of pure theoretical interest, arises naturally in many contexts: designing privacy-preserving social-networks, private peer-to-peer computations, vehicle-to-vehicle
networks and the “internet of things” are some of the examples.
In this paper, we initiate the study of “topology-hiding computation” in the computational setting. We give formal definitions in both simulation-based and indistinguishability-based flavors. We
show that, even for fail-stop adversaries, there are some strong impossibility results. Despite this, we show that protocols for topology-hiding computation can be constructed in the semi-honest
and fail-stop models, if we somewhat restrict the set of nodes the adversary may corrupt.
• Giulia Alberini, Tal Moran and Alon Rosen
Public Verification of Private Effort
TCC 2015 [BibTex]
author = {Giulia Alberini and Tal Moran and Alon Rosen},
title = {Public Verification of Private Effort},
editor = {Yevgeniy Dodis and Jesper Buus Nielsen},
booktitle = {TCC 2015},
pages = {159--181},
series = {Lecture Notes in Computer Science},
volume = {9014},
year = {2015},
publisher = {Springer},
ee = {\url{https://eprint.iacr.org/2014/983}},
We introduce a new framework for polling responses from a large population. Our framework allows gathering information without violating the responders' anonymity and at the same time enables
public verification of the poll's result. In contrast to prior approaches to the problem, we do not require trusting the pollster for faithfully announcing the poll's results, nor do we rely on
strong identity verification.
We propose an “effort based” polling protocol whose results can be publicly verified by constructing a “responder certification graph” whose nodes are labeled by responders' replies to the poll,
and whose edges cross-certify that adjacent nodes correspond to honest participants. Cross-certification is achieved using a newly introduced (privately verifiable) “Private Proof of Effort”
(PPE). In effect, our protocol gives a general method for converting privately-verifiable proofs into a publicly-verifiable protocol. The soundness of the transformation relies on expansion
properties of the certification graph.
Our results are applicable to a variety of settings in which crowd-sourced information gathering is required. This includes crypto-currencies, political polling, elections, recommendation
systems, viewer voting in TV shows, and prediction markets.
• Ranjit Kumaresan, Tal Moran and Iddo Bentov
How to Use Bitcoin to Play Decentralized Poker
CCS 2015 [BibTex]
author = {Ranjit Kumaresan and Tal Moran and Iddo Bentov},
title = {How to Use Bitcoin to Play Decentralized Poker},
booktitle = {CCS 2015},
pages = {195--206},
year = {2015},
url = {http://doi.acm.org/10.1145/2810103.2813712},
doi = {10.1145/2810103.2813712},
Back and Bentov (arXiv 2014) and Andrychowicz et al. (Security and Privacy 2014) introduced techniques to perform secure multiparty computations on Bitcoin. Among other things, these works
constructed lottery protocols that ensure that any party that aborts after learning the outcome pays a monetary penalty to all other parties. Following this, Andrychowicz et al. (Bitcoin Workshop
2014) and concurrently Bentov and Kumaresan (Crypto 2014) extended the solution to arbitrary secure function evaluation while guaranteeing fairness in the following sense: any party that aborts
after learning the output pays a monetary penalty to all parties that did not learn the output. Andrychowicz et al. (Bitcoin Workshop 2014) also suggested extending to scenarios where parties
receive a payoff according to the output of a secure function evaluation, and outlined a 2-party protocol for the same that in addition satisfies the notion of fairness described above.
In this work, we formalize, generalize, and construct multiparty protocols for the primitive suggested by Andrychowicz et al. We call this primitive secure cash distribution with penalties. Our
formulation of secure cash distribution with penalties poses it as a multistage reactive functionality (i.e., more general than secure function evaluation) that provides a way to securely
implement smart contracts in a decentralized setting, and consequently suffices to capture a wide variety of stateful computations involving data and/or money, such as {decentralized} auctions,
markets, and games such as poker, etc. Our protocol realizing secure cash distribution with penalties works in a hybrid model where parties have access to a claim-or-refund transaction
functionality $\mathcal{F}_{\mathrm{CR}}^\star$ which can be efficiently realized in (a variant of) Bitcoin, and is otherwise independent of the Bitcoin ecosystem. We emphasize that our protocol
is dropout-tolerant in the sense that any party that drops out during the protocol is forced to pay a monetary penalty to all other parties. Our formalization and construction generalize both
secure computation with penalties of Bentov and Kumaresan (Crypto 2014),and secure lottery with penalties of Andrychowicz et al.\ (Security and Privacy 2014).
• Ilan Komargodski, Tal Moran, Moni Naor, Rafael Pass, Alon Rosen and Eylon Yogev
One-Way Functions and (Im)perfect Obfuscation
FOCS 2014 [BibTex]
author = {Ilan Komargodski and Tal Moran and Moni Naor and Rafael Pass and Alon Rosen and Eylon Yogev},
title = {One-Way Functions and (Im)perfect Obfuscation},
booktitle = {FOCS 2014},
pages = {374--383},
year = {2014},
ee = {http://dx.doi.org/10.1109/FOCS.2014.47},
A program obfuscator takes a program and outputs an "scrambled" version of it, where the goal is that the obfuscated program will not reveal much about its structure beyond what is apparent from
executing it. There are several ways of formalizing this goal. Specifically, in indistinguishability obfuscation, first defined by Barak et al. (CRYPTO 2001), the requirement is that the results
of obfuscating any two functionally equivalent programs (circuits) will be computationally indistinguishable. Recently, a fascinating candidate construction for indistinguishability obfuscation
was proposed by Garg et al. (FOCS 2013). This has led to a flurry of discovery of intriguing constructions of primitives and protocols whose existence was not previously known (for instance,
fully deniable encryption by Sahai and Waters, STOC 2014). Most of them explicitly rely on additional hardness assumptions, such as one-way functions.
Our goal is to get rid of this extra assumption. We cannot argue that indistinguishability obfuscation of all polynomial-time circuits implies the existence of one-way functions, since if $P =
NP$, then program obfuscation (under the indistinguishability notion) is possible. Instead, the ultimate goal is to argue that if $P \neq NP$ and program obfuscation is possible, then one-way
functions exist.
Our main result is that if $NP \not\subseteq ioBPP$ and there is an efficient (even imperfect) indistinguishability obfuscator, then there are one-way functions. In addition, we show that the
existence of an indistinguishability obfuscator implies (unconditionally) the existence of SZK-arguments for $NP$. This, in turn, provides an alternative version of our main result, based on the
assumption of hard-on-the average $NP$ problems. To get some of our results we need obfuscators for simple programs such as 3CNF formulas.
• Mohammad Mahmoody, Tal Moran and Salil Vadhan
Publicly Verifiable Proofs of Sequential Work
ITCS 2013 [BibTex]
author = {Mohammad Mahmoody and Tal Moran and Salil Vadhan},
title = {Publicly Verifiable Proofs of Sequential Work},
editor = {Robert D. Kleinberg},
booktitle = {ITCS 2013},
pages = {373--388},
year = {2013},
month = {January},
publisher = {ACM},
We construct a publicly verifiable protocol for proving computational work based on collision-resistant hash functions and a new plausible complexity assumption regarding the existence of
“inherently sequential” hash functions. Our protocol is based on a novel construction of time-lock puzzles. Given a sampled “puzzle” $\mathcal{P}\mathbin{\stackrel{\tiny{\$}}{\gets}} \mathbf{D}
_n$, where $n$ is the security parameter and $\mathbf{D}_n$ is the distribution of the puzzles, a corresponding “solution” can be generated using $N$ evaluations of the sequential hash function,
where $N>n$ is another parameter, while any feasible adversarial strategy for generating valid solutions must take at least as much time as $\Omega(N)$ sequential evaluations of the hash function
after receiving $\mathcal{P}$. Thus, valid solutions constitute a “proof” that $\Omega(N)$ parallel time elapsed since $\mathcal{P}$ was received. Solutions can be publicly and efficiently
verified in time $poly(n) \cdot polylog(N)$. Applications of these “time-lock puzzles” include noninteractive timestamping of documents (when the distribution over the possible documents
corresponds to the puzzle distribution $\mathbf{D}_n$) and universally verifiable CPU benchmarks.
Our construction is secure in the standard model under complexity assumptions (collision-resistant hash functions and inherently sequential hash functions), and makes black-box use of the
underlying primitives. Consequently, the corresponding construction in the random oracle model is secure unconditionally. Moreover, as it is a public-coin protocol, it can be made non-interactive
in the random oracle model using the Fiat-Shamir Heuristic.
Our construction makes a novel use of “depth-robust” directed acyclic graphs—ones whose depth remains large even after removing a constant fraction of vertices—which were previously studied for
the purpose of complexity lower bounds. The construction bypasses a recent negative result of Mahmoody, Moran, and Vadhan (CRYPTO `11) for time-lock puzzles in the random oracle model, which
showed that it is impossible to have time-lock puzzles like ours in the random oracle model if the puzzle generator also computes a solution together with the puzzle.
• Shahram Khazaei, Tal Moran and Douglas Wikström
A Mix-Net From Any CCA2 Secure Cryptosystem
Asiacrypt 2012 [BibTex]
author = {Shahram Khazaei and Tal Moran and Douglas Wikstr\"{o}m},
title = {A Mix-Net From Any CCA2 Secure Cryptosystem},
editor = {Xiaoyun Wang and Kazue Sako},
booktitle = {Asiacrypt 2012},
pages = {607--625},
series = {Lecture Notes in Computer Science},
volume = {7658},
year = {2012},
month = {December},
publisher = {Springer},
We construct a provably secure mix-net from any CCA2 secure cryptosystem. The mix-net is secure against active adversaries that statically corrupt less than $\lambda$ out of $k$ mix-servers,
where $\lambda$ is a threshold parameter, and it is robust provided that at most $\min(\lambda-1,k-\lambda)$ mix-servers are corrupted.
The main component of our construction is a mix-net that outputs the correct result if all mix-servers behaved honestly, and aborts with probability $1-O(H^{-(t-1)})$ otherwise (without
disclosing anything about the inputs), where $t$ is an auxiliary security parameter and $H$ is the number of honest parties. The running time of this protocol for long messages is roughly $3t c$,
where $c$ is the running time of Chaum's mix-net (1981).
• Yevgeniy Dodis, Abhishek Jain, Tal Moran and Daniel Wichs
Counterexamples to Hardness Amplification Beyond Negligible
TCC 2012 [BibTex]
author = {Yevgeniy Dodis and Abhishek Jain and Tal Moran and Daniel Wichs},
title = {Counterexamples to Hardness Amplification Beyond Negligible},
booktitle = {TCC 2012},
pages = {476--493},
series = {Lecture Notes in Computer Science},
volume = {7194},
year = {2012},
month = {March},
publisher = {Springer},
isbn = {978-3-642-28914-9},
If we have a problem that is mildly hard, can we create a problem that is significantly harder? A natural approach to hardness amplification is the “direct product”; instead of asking an attacker
to solve a single instance of a problem, we ask the attacker to solve several independently generated ones. Interestingly, proving that the direct product amplifies hardness is often highly
non-trivial, and in some cases may be false. For example, it is known that the direct product (i.e. “parallel repetition”) of general interactive games may not amplify hardness at all. On the
other hand, positive results show that the direct product does amplify hardness for many basic primitives such as one-way functions/relations, weakly-verifiable puzzles, and signatures.
Even when positive direct product theorems are shown to hold for some primitive, the parameters are surprisingly weaker than what we may have expected. For example, if we start with a weak
one-way function that no poly-time attacker can break with probability $> \frac{1}{2}$, then the direct product provably amplifies hardness to some negligible probability. Naturally, we would
expect that we can amplify hardness exponentially, all the way to $2^{-n}$ probability, or at least to some fixed/known negligible such as $n^{-\log n}$ in the security parameter $n$, just by
taking sufficiently many instances of the weak primitive. Although it is known that such parameters cannot be proven via black-box reductions, they may seem like reasonable conjectures, and, to
the best of our knowledge, are widely believed to hold. In fact, a conjecture along these lines was introduced in a survey of Goldreich, Nisan and Wigderson (ECCC '95). In this work, we show that
such conjectures are false by providing simple but surprising counterexamples. In particular, we construct weakly secure signatures and one-way functions, for which standard hardness
amplification results are known to hold, but for which hardness does not amplify beyond just negligible. That is, for any negligible function $\epsilon(n)$, we instantiate these primitives so
that the direct product can always be broken with probability $\epsilon(n)$, no matter how many copies we take.
• Mohammad Mahmoody, Tal Moran and Salil Vadhan
Time-Lock Puzzles in the Random Oracle Model
CRYTPO 2011 [BibTex]
author = {Mohammad Mahmoody and Tal Moran and Salil Vadhan},
title = {Time-Lock Puzzles in the Random Oracle Model},
booktitle = {CRYTPO 2011},
pages = {39--50},
series = {Lecture Notes in Computer Science},
volume = {6841},
year = {2011},
month = {August},
publisher = {Springer-Verlag},
A time-lock puzzle is a mechanism for sending messages “to the future”. The sender publishes a puzzle whose solution is the message to be sent, thus hiding it until enough time has elapsed for
the puzzle to be solved. For time-lock puzzles to be useful, generating a puzzle should take less time than solving it. Since adversaries may have access to many more computers than honest
solvers, massively parallel solvers should not be able to produce a solution much faster than serial ones.
To date, we know of only one mechanism that is believed to satisfy these properties: the one proposed by Rivest, Shamir and Wagner (1996), who originally introduced the notion of time-lock
puzzles. Their puzzle is based on the serial nature of exponentiation and the hardness of factoring, and is therefore vulnerable to advances in factoring techniques (as well as to quantum
In this work, we study the possibility of constructing time-lock puzzles in the random-oracle model. Our main result is negative, ruling out time-lock puzzles that require more parallel time to
solve than the total work required to generate a puzzle. In particular, this rules out black-box constructions of such time-lock puzzles from one-way permutations and collision-resistant
hash-functions. On the positive side, we construct a time-lock puzzle with a linear gap in parallel time: a new puzzle can be generated with one round of $n$ parallel queries to the random
oracle, but $n$ rounds of serial queries are required to solve it (even for massively parallel adversaries).
• John Kelsey, Andrew Regenscheid, Tal Moran and David Chaum
Attacking Paper-Based E2E Voting Systems
Towards Trustworthy Elections [BibTex]
author = {John Kelsey and Andrew Regenscheid and Tal Moran and David Chaum},
title = {Attacking Paper-Based E2E Voting Systems},
booktitle = {Towards Trustworthy Elections},
pages = {370--387},
series = {Lecture Notes in Computer Science},
volume = {6000},
year = {2010},
month = {August},
publisher = {Springer},
isbn = {978-3-642-12979-7},
In this paper, we develop methods for constructing vote-buying/coercion attacks on end-to-end voting systems, and describe vote-buying/coercion attacks on three proposed end-to-end voting
systems: Punchscan, Pret-a-voter and ThreeBallot. We also demonstrate a different attack on Punchscan, which could permit corrupt election officials to change votes without detection in some
cases. Additionally, we consider some generic attacks on end-to-end voting systems.
• Tal Moran and Tyler Moore
The Phish-Market Protocol: Secure Sharing Between Competitors
IEEE Security & Privacy, 8(4):40–45, 2010 (A version of this article aimed at a more technical audience was previously published in FC 2010.) [BibTex]
author = {Tal Moran and Tyler Moore},
title = {The Phish-Market Protocol: Secure Sharing Between Competitors},
pages = {40--45},
volume = {8},
number = {4},
year = {2010},
month = {July},
note = {A version of this article aimed at a more technical audience was previously published in FC 2010.},
journal = {IEEE Security \& Privacy},
doi = {10.1109/MSP.2010.138},
The Phish-Market protocol encourages take-down companies to share information about malicious websites by compensating them for this data without revealing sensitive information to their
competitors. Cryptography lets contributing firms verify payment amounts without learning which offered website URLs were “purchased.”
• Tal Moran and Moni Naor
Basing Cryptographic Protocols on Tamper-Evident Seals
Theoretical Computer Science, 411:1283–1310, 2010 [BibTex]
author = {Tal Moran and Moni Naor},
title = {Basing Cryptographic Protocols on Tamper-Evident Seals},
pages = {1283--1310},
volume = {411},
year = {2010},
month = {March},
publisher = {Elsevier Science Publishers Ltd.},
journal = {Theoretical Computer Science},
doi = {10.1016/j.tcs.2009.10.023},
In this paper we attempt to formally study two very intuitive physical models: sealed envelopes and locked boxes, often used as illustrations for common cryptographic operations. We relax the
security properties usually required from locked boxes (such as in bit-commitment protocols) and require only that a broken lock or torn envelope be identifiable to the original sender. Unlike
the completely impregnable locked box, this functionality may be achievable in real life, where containers having this property are called “tamper-evident seals”. Another physical object with
this property is the “scratch-off card”, often used in lottery tickets. We consider three variations of tamper-evident seals, and show that under some conditions they can be used to implement
oblivious transfer, bit-commitment and coin flipping. We also show a separation between the three models. Of particular interest, we give a strongly-fair coin flipping protocol with bias bounded
by $O(1/r)$ (where r is the number of rounds), beating the best known bias in the standard model even with cryptographic assumptions.
[Preliminary version appeared in ICALP 2005 |
author = {Tal Moran and Moni Naor},
title = {Basing Cryptographic Protocols on Tamper-Evident Seals},
editor = {L. Caires et al.},
booktitle = {ICALP 2005},
pages = {285--297},
series = {Lecture Notes in Computer Science},
volume = {3580},
year = {2005},
month = {Jul},
publisher = {Springer-Verlag},
• Tal Moran and Moni Naor
Split-Ballot Voting: Everlasting Privacy With Distributed Trust
ACM Transactions on Information and System Security, 13:16:1–16:43, 2010 [BibTex]
author = {Tal Moran and Moni Naor},
title = {Split-Ballot Voting: Everlasting Privacy With Distributed Trust},
pages = {16:1--16:43},
volume = {13},
year = {2010},
month = {March},
publisher = {ACM},
journal = {ACM Transactions on Information and System Security},
doi = {http://doi.acm.org/10.1145/1698750.1698756},
In this paper we propose a new voting protocol with desirable security properties. The voting stage of the protocol can be performed by humans without computers; it provides every voter with the
means to verify that all the votes were counted correctly (universal verifiability) while preserving ballot secrecy. The protocol has “everlasting privacy”: even a computationally unbounded
adversary gains no information about specific votes from observing the protocol's output. Unlike previous protocols with these properties, this protocol distributes trust between two authorities:
a single corrupt authority will not cause voter privacy to be breached. Finally, the protocol is receipt-free: a voter cannot prove how she voted even she wants to do so. We formally prove the
security of the protocol in the Universal Composability framework, based on number-theoretic assumptions.
[Preliminary version appeared in CCS 2007 |
author = {Tal Moran and Moni Naor},
title = {Split-Ballot Voting: Everlasting Privacy With Distributed Trust},
booktitle = {CCS 2007},
pages = {246--255},
year = {2007},
month = {October},
publisher = {ACM},
isbn = {978-1-59593-703-2},
• Dov Gordon, Yuval Ishai, Tal Moran, Rafail Ostrovsky and Amit Sahai
On Complete Primitives for Fairness
TCC 2010 [BibTex]
author = {Dov Gordon and Yuval Ishai and Tal Moran and Rafail Ostrovsky and Amit Sahai},
title = {On Complete Primitives for Fairness},
editor = {Daniele Micciancio},
booktitle = {TCC 2010},
pages = {91--108},
series = {Lecture Notes in Computer Science},
volume = {5978},
year = {2010},
month = {February},
publisher = {Springer Berlin / Heidelberg},
For secure two-party and multi-party computation with abort, classification of which primitives are complete has been extensively studied in the literature. However, for fair secure computation,
where (roughly speaking) either all parties learn the output or none do, the question of complete primitives has remained largely unstudied. In this work, we initiate a rigorous study of
completeness for primitives that allow fair computation. We show the following results:
□ No “short” primitive is complete for fairness. In surprising contrast to other notions of security for secure two-party computation, we show that for fair secure two-party computation, no
primitive of size $O(\log k)$ is complete, where $k$ is a security parameter. This is the case even if we can enforce parallelism in calls to the primitives (i.e., the adversary does not get
output from any primitive in a parallel call until it sends input to all of them). This negative result holds regardless of any computational assumptions.
□ Coin Flipping and Simultaneous Broadcast are not complete for fairness. The above result rules out the completeness of two natural candidates: coin flipping (for any number of coins) and
simultaneous broadcast (for messages of arbitrary length).
□ A fairness hierarchy. We clarify the fairness landscape further by exhibiting the existence of a “fairness hierarchy”. We show that for every “short” $\ell = O(\log k)$, no protocol making
(serial) access to any $\ell$-bit primitive can be used to construct even a $(\ell+1)$-bit simultaneous broadcast.
□ Positive results. To complement the negative results, we exhibit a $k$-bit primitive that is complete for two-party fair secure computation. This primitive implements a “fair reconstruction”
procedure for a secret sharing scheme with some robustness properties. We show how to generalize this result to the multi-party setting.
□ Fairness combiners. We also introduce the question of constructing a protocol for fair secure computation from primitives that may be faulty. We show a simple functionality that is complete
for two-party fair computation when the majority of its instances are honest. On the flip side, we show that this result is tight: no functionality is complete for fairness if half (or more)
of the instances can be malicious.
• Tal Moran and Tyler Moore
The Phish Market Protocol: Securely Sharing Attack Data Between Competitors
FC 2010 [BibTex]
author = {Tal Moran and Tyler Moore},
title = {The Phish Market Protocol: Securely Sharing Attack Data Between Competitors},
editor = {Radu Sion},
booktitle = {FC 2010},
pages = {222--237},
series = {Lecture Notes in Computer Science},
volume = {6052},
year = {2010},
month = {January},
A key way in which banks mitigate the effects of phishing is to remove fraudulent websites or suspend abusive domain names. This `take-down' is often subcontracted to specialist firms. Prior work
has shown that these take-down companies refuse to share `feeds' of phishing website URLs with each other, and consequently, many phishing websites are not removed because the firm with the
take-down contract remains unaware of their existence. The take-down companies are reticent to exchange feeds, fearing that competitors with less comprehensive lists might `free-ride' off their
efforts by not investing resources to find new websites, as well as use the feeds to poach clients. In this paper, we propose the Phish Market protocol, which enables companies with less
comprehensive feeds to learn about websites impersonating their own clients that are held by other firms. The protocol is designed so that the contributing firm is compensated only for those
websites affecting its competitor's clients and only those previously unknown to the receiving firm. Crucially, the protocol does not reveal to the contributing firm which URLs are needed by the
receiver, as this is viewed as sensitive information by take-down firms. Using complete lists of phishing URLs obtained from two large take-down companies, our elliptic-curve-based implementation
added a negligible average 5 second delay to securely share URLs.
• Tal Moran, Moni Naor and Gil Segev
Deterministic History-Independent Strategies for Storing Information on Write-Once Memories
Theory of Computing, 5(1):43–67, 2009 [BibTex]
author = {Tal Moran and Moni Naor and Gil Segev},
title = {Deterministic History-Independent Strategies for Storing Information on Write-Once Memories},
pages = {43--67},
volume = {5},
number = {1},
year = {2009},
publisher = {Theory of Computing},
url = {http://www.theoryofcomputing.org/articles/v005a002},
journal = {Theory of Computing},
doi = {10.4086/toc.2009.v005a002},
Motivated by the challenging task of designing “secure” vote storage mechanisms, we deal with information storage mechanisms that operate in extremely hostile environments. In such environments,
the majority of existing techniques for information storage and for security are susceptible to powerful adversarial attacks. In this setting, we propose a mechanism for storing a set of at most
K elements from a large universe of size N on write-once memories in a manner that does not reveal the insertion order of the elements. We consider a standard model for write-once memories, in
which the memory is initialized to the all 0's state, and the only operation allowed is flipping bits from 0 to 1. Whereas previously known constructions were either inefficient (required $\Theta
(K^2)$ memory), randomized, or employed cryptographic techniques which are unlikely to be available in hostile environments, we eliminate each of these undesirable properties. The total amount of
memory used by the mechanism is linear in the number of stored elements and poly-logarithmic in the size of the universe of elements. In addition, we consider one of the classical distributed
computing problems: conflict resolution in multiple-access channels. By establishing a tight connection with the basic building block of our mechanism, we construct the first deterministic and
non-adaptive conflict resolution algorithm whose running time is optimal up to poly-logarithmic factors.
[Preliminary version appeared in ICALP 2007 |
author = {Tal Moran and Moni Naor and Gil Segev},
title = {Deterministic History-Independent Strategies for Storing Information on Write-Once Memories},
booktitle = {ICALP 2007},
series = {Lecture Notes in Computer Science},
volume = {4596},
year = {2007},
month = {July},
publisher = {Springer},
isbn = {987-5-540-73419-2},
• Josh Benaloh, Tal Moran, Lee Naish, Kim Ramchen and Vanessa Teague
Shuffle-Sum: Coercion-Resistant Verifiable Tallying for STV Voting
IEEE Transactions on Information Forensics and Security, 4(4):685–698, 2009 [BibTex]
author = {Josh Benaloh and Tal Moran and Lee Naish and Kim Ramchen and Vanessa Teague},
title = {Shuffle-Sum: Coercion-Resistant Verifiable Tallying for STV Voting},
pages = {685--698},
volume = {4},
number = {4},
year = {2009},
month = {December},
journal = {IEEE Transactions on Information Forensics and Security},
doi = {10.1109/TIFS.2009.2033757},
There are many advantages to voting schemes in which voters rank all candidates in order, rather than just choosing their favourite. However, these schemes inherently suffer from a coercion
problem when there are many candidates, because a coercer can demand a certain permutation from a voter and then check whether that permutation appears during tallying. Recently developed
cryptographic voting protocols allow anyone to audit an election (universal verifiability), but existing systems are either not applicable to ranked voting at all, or reveal enough information
about the ballots to make voter coercion possible. We solve this problem for the popular single transferable vote (STV) ranked voting system, by constructing an algorithm for the verifiable
tallying of encrypted votes. Our construction improves upon existing work because it extends to multiple-seat STV and reveals less information than other schemes. The protocol is based on
verifiable shuffling of homomorphic encryptions, a well-studied primitive in the voting arena. Our protocol is efficient enough to be practical, even for a large election.
• Tal Moran, Ronen Shaltiel and Amnon Ta-Shma
Non-interactive Timestamping in the Bounded Storage Model
J. Cryptology, 22(2):189–226, 2009 [BibTex]
author = {Tal Moran and Ronen Shaltiel and Amnon Ta-Shma},
title = {Non-interactive Timestamping in the Bounded Storage Model},
pages = {189--226},
volume = {22},
number = {2},
year = {2009},
month = {April},
journal = {J. Cryptology},
doi = {10.1007/s00145-008-9035-9},
A timestamping scheme is a mechanism allowing one party, the “stamper”, to prove that it knew of a certain document at some earlier time. We say that such a scheme is passive if a stamper can
stamp a document without communicating with any other player. The only communication performed is at validation time. Passive timestamping has many advantages, such as information theoretic
privacy and enhanced robustness. Passive timestamping, however, is not possible against polynomial time adversaries that have unbounded (but polynomial) storage at their disposal. As a result, no
passive timestamping schemes were constructed up to date. We show that passive timestamping is possible in the Bounded Storage Model. I.e., where there is an upper bound on the amount of storage
that the adversary has and all players have access to a long random string. To the best of our knowledge, this is the first example of a cryptographic task that is possible in the bounded storage
model, but is impossible in the “standard cryptographic setting”, even assuming cryptographic assumptions. We give an explicit construction that is secure against all bounded storage adversaries,
and a significantly more efficient construction secure against all bounded storage adversaries that run in polynomial time.
[Preliminary version appeared in CRYPTO 2004 |
author = {Tal Moran and Ronen Shaltiel and Amnon Ta-Shma},
title = {Non-interactive Timestamping in the Bounded Storage Model.},
editor = {Matthew K. Franklin},
booktitle = {CRYPTO 2004},
pages = {460--476},
series = {Lecture Notes in Computer Science},
volume = {3152},
year = {2004},
month = {August},
publisher = {Springer},
• Tal Moran and Gil Segev
David and Goliath Commitments: UC Computation for Asymmetric Parties Using Tamper-Proof Hardware
Eurocrypt 2008 [BibTex]
author = {Tal Moran and Gil Segev},
title = {David and Goliath Commitments: UC Computation for Asymmetric Parties Using Tamper-Proof Hardware},
booktitle = {Eurocrypt 2008},
pages = {527--544},
series = {Lecture Notes in Computer Science},
volume = {4965},
year = {2008},
month = {April},
publisher = {Springer},
isbn = {978-3-540-78966-6},
Designing secure protocols in the Universal Composability (UC) framework confers many advantages. In particular, it allows the protocols to be securely used as building blocks in more complex
protocols, and assists in understanding their security properties. Unfortunately, most existing models in which universally composable computation is possible (for useful functionalities) require
a trusted setup stage. Recently, Katz [Eurocrypt '07] proposed an alternative to the trusted setup assumption: tamper-proof hardware. Instead of trusting a third party to correctly generate the
setup information, each party can create its own hardware tokens, which it sends to the other parties. Each party is only required to trust that its own tokens are tamper-proof. Katz designed a
UC commitment protocol that requires both parties to generate hardware tokens. In addition, his protocol relies on a specific number-theoretic assumption. In this paper, we construct UC
commitment protocols for “David” and “Goliath”: we only require a single party (Goliath) to be capable of generating tokens. We construct a version of the protocol that is secure for
computationally unbounded parties, and a more efficient version that makes computational assumptions only about David (we require only the existence of a one-way function). Our protocols are
simple enough to be performed by hand on David's side. These properties may allow such protocols to be used in situations which are inherently asymmetric in real-life, especially those involving
individuals versus large organizations. Classic examples include voting protocols (voters versus “the government”) and protocols involving private medical data (patients versus insurance-agencies
or hospitals).
• Tal Moran and Moni Naor
Receipt-Free Universally-Verifiable Voting With Everlasting Privacy
CRYPTO 2006 [BibTex]
author = {Tal Moran and Moni Naor},
title = {Receipt-Free Universally-Verifiable Voting With Everlasting Privacy},
editor = {Cynthia Dwork},
booktitle = {CRYPTO 2006},
pages = {373--392},
series = {Lecture Notes in Computer Science},
volume = {4117},
year = {2006},
month = {September},
publisher = {Springer},
isbn = {3-540-37432-9},
We present the first universally verifiable voting scheme that can be based on a general assumption (existence of a non-interactive commitment scheme). Our scheme is also the first receipt-free
scheme to give “everlasting privacy” for votes: even a computationally unbounded party does not gain any information about individual votes (other than what can be inferred from the final tally).
Our voting protocols are designed to be used in a “traditional” setting, in which voters cast their ballots in a private polling booth (which we model as an untappable channel between the voter
and the tallying authority). Following in the footsteps of Chaum and Neff, our protocol ensures that the integrity of an election cannot be compromised even if the computers running it are all
corrupt (although ballot secrecy may be violated in this case). We give a generic voting protocol which we prove to be secure in the Universal Composability model, given that the underlying
commitment is universally composable. We also propose a concrete implementation, based on the hardness of discrete log, that is more efficient.
• Tal Moran and Moni Naor
Polling with Physical Envelopes: A Rigorous Analysis of a Human-Centric Protocol
Eurocrypt 2006 [BibTex]
author = {Tal Moran and Moni Naor},
title = {Polling with Physical Envelopes: A Rigorous Analysis of a Human-Centric Protocol},
editor = {Serge Vaudenay},
booktitle = {Eurocrypt 2006},
pages = {88--108},
series = {Lecture Notes in Computer Science},
volume = {4004},
year = {2006},
month = {May},
publisher = {Springer-Verlag},
We propose simple, realistic protocols for polling that allow the responder to plausibly repudiate his response, while at the same time allow accurate statistical analysis of poll results. The
protocols use simple physical objects (envelopes or scratch-off cards) and can be performed without the aid of computers. One of the main innovations of this work is the use of techniques from
theoretical cryptography to rigorously prove the security of a realistic, physical protocol. We show that, given a few properties of physical envelopes, the protocols are unconditionally secure
in the universal composability framework.
• A great example of "human cryptography" (i.e., cryptographic protocols that can be performed with materials in every kitchen) is this puzzle, due to Naor, Naor and Reingold, and published in the
highly respected Journal of Craptology | {"url":"https://talmoran.net/","timestamp":"2024-11-05T16:02:04Z","content_type":"text/html","content_length":"108632","record_id":"<urn:uuid:9e08656c-ba93-4fcb-a4b6-7bb786ce8b28>","cc-path":"CC-MAIN-2024-46/segments/1730477027884.62/warc/CC-MAIN-20241105145721-20241105175721-00286.warc.gz"} |
how to calculate critical speed of ball mill
A Slice Mill is the same diameter as the production mill but shorter in length. Request Price Quote. Click to request a ball mill quote online or call to speak with an expert at Paul O. Abbe® to
help you determine which design and size ball mill would be best for your process. See our Size Reduction Options.
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u 2 − fresh ore feed rate, u 3 − mill critical speed fraction, u 4 − sump dilution water . ... Ball mills can grind a wide range of materials, including metals, ceramics, and polymers, and ...
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A ball mill, a type of grinder, is a cylindrical device used in grinding (or mixing) materials like ores, chemicals, ceramic raw materials and paints. Ball mills rotate around a horizontal axis,
partially filled with the material to be ground plus the grinding medium. Different materials are used as media, including ceramic balls, flint pebbles ...
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Similarly, from Fig., it would appear that the throughput varies directly with the speed of rotation, for speeds up to about 70% of the critical. At some point above this speed the curve would be
expected to become horizontal, since at the critical speed the charge would be spread around the mill and horizontal flow would then be unlikely.
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how to calculate cement mill critical speed . 04116 Where : Nc = Critical speed, CEMENT MILL FORMULAS MILL CRITICAL VELOCITY = 76 / (D)^1/2 MILL L ball . cement ball mill critical speed Crusher|
Granite Crusher .
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https:// Learn about Ball Mill Critical Speed and its effect on inner charge movements. The effect of Ball Mill RPM s...
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What should the opening speed of a ball mill be the critical speed? The opening speed of a ball mill should not be equal to the critical speed. Operating a mill at critical speed can lead to
excessive wear and vibration. The opening speed is typically lower than critical speed.
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The critical speed of the mill, c, is defined as the speed at which a single ball will just remain against the wall for a full cycle. At the top of the cycle =0 and Fc Fg () mp 2 cDm 2 mpg () c
2g Dm 1/2 () The critical speed is usually expressed in terms of the number of revolutions per second Nc c 2 1 2 2g Dm 1/2 (2×)1/2 ...
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pirical formulae for wet rod mills and wet and dry ball mills, where the diameter occurs as The formula contains in accordance with above explana tion the factor Os (fraction of critical speed)
and the filling rate of the grind ing media charge. It is important to stress, however, that a simultaneous slippagefree mill
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online live calculators for grinding calculations, Ball mill, tube mill, critical speed, Degree of filling balls, Arm of gravity, mill net and gross power. Optimization; Online Training; ...
Critical Speed (nc) Mill Speed (n) Please Enter / Stepto Input Values Mill Eff. Dia Deff, m CALCULATED VALUE OF nc, rpm ...
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Critical speed (in rpm) = /sqrt(D d) with D the diameter of the mill in meters and d the diameter of the largest grinding ball you will use for experiment (also expressed in meters)
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To compute for shaft power | ball mill length, six essential parameters are needed and these parameters are Value of C, Volume Load Percentage (J), % Critical Speed (Vcr), Bulk Density (), Mill
Length (L) and Mill Internal Diameter (D). The formula for calculating shaft power | ball mill length: P = x C X J X V cr x (1 ) x [1 ...
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Prompt : Caesar is a famous mining equipment manufacturer wellknown both at home and abroad, major in producing stone crushing equipment, mineral separation equipment, limestone grinding
equipment, etc. formula for critical speed of ball mill YouTube . 15 Oct 2013 ... critical speed tumbling mill formula.
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The optimum speed varies as a percentage of the critical speed depending on the viscosity of the material being ground. For the dry powders used in pyrotechnics, the optimum speed will be 65% of
the critical speed. Using the interactive calculator on this page will help you determine the optimal speed for your mill without having to reach for a ...
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Rowling are used to calculate the mill power draw. The Morgärdshammar equation and the IMM equations are shown for comparison. The method of use is similar to the AM section Ball Mill Design The
ball mill designs also follow the Bond/Rowlings method with comparison with other methods. Again the method of use is the same
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The critical speed n (rpm) when the balls are attached to the wall due to centrifugation: Figure Displacement of balls in mill. ... The company clams this new ball mill will be helpful to enable
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In most cases, the ideal mill speed will have the media tumbling from the top of the pile (the shoulder) to the bottom (the toe) with many impacts along the way. The ideal mill speed is usually
somewhere between 55% to 75% of critical speed. Critical Mill Speed. Critical Speed (left) is the speed at which the outer layer of media centrifuges ...
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Critical Speed = / √( ) Critical Speed ≈ RPM Therefore, the critical speed of the ball mill is approximately revolutions per minute.
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Generally, the smaller diameter mills operate at a higher percentage of critical speed than the larger mills. Grinding Mill Horse Power. ... usually 43° for dry grinding slow speed ball mill, 51°
for normal ball mill speeds. *The theoretical exponent is, but actual results indicate that is more nearly correct. ...
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These dimensions result in a critical rotational speed of 130 rpm. The actual rotational speed used was 90 rpm, 70% of the critical value, as suggested by [23]. The apparent density, ρ a, and ...
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Often, this minimum charge volume is roughly doubled (⅔·55 = 37% V; 40% if rounded to the nearest ten), so that a reasonable charge layer covers the balls. If so, half of the occupied volume ...
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Calculate the critical speed, in revolutions per minute, for a ball mill 1600 mm in diameter charged with 100 mm balls..... a) 149. b) 276. c) 321
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According to available literature, the operating speeds of AG mills are much higher than conventional tumbling mills and are in the range of 8085% of the critical speed. SAG mills of comparable
size but containing say 10% ball charge (in addition to the rocks), normally, operate between 70 and 75% of the critical speed.
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Objectives. At the end of this lesson students should be able to: Explain the grinding process. Distinguish between crushing and grinding. Compare and contrast different type of equipment and
their components used for grinding. Identify key variables for process control. Design features of grinding equipment (SAG, BALL and ROD MILLS)
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The diameter of the balls used in ball mills play a significant role in the improvement and optimization of the efficiency of the mill [4]. The optimum rotation speed of a mill, which is the
speed at which optimum size reduction takes place, is determined in terms of the percentage of the critical speed of the mill [8]. The critical speed is
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The mill was rotated at 50, 62, 75 and 90% of the critical speed. Six lifter bars of rectangular crosssection were used at equal spacing. The overall motion of the balls at the end of five
revolutions is shown in Figure 4. As can be seen from the figure, the overall motion of the balls changes with the mill speed inasmuch as the shoulder ...
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The formula for calculating shaft diameter: D = ³√ ( x 106 x P / N) Where: D = Shaft Diameter. P = Shaft Power. N = 75% of Critical Speed. Let's solve an example; Find the shaft diameter when the
shaft power is 28 and the 75% of critical speed is 7. This implies that;
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To use this online calculator for Critical Speed of Conical Ball Mill, enter Radius of Ball Mill (R) Radius of Ball (r) and hit the calculate button. Here is how the Critical Speed of Conical
Ball Mill calculation can be explained with given input values > = 1/(2*pi)*sqrt( [g]/()).
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Speed rate refers to the ratio of the speed of the mill to the critical speed, where the critical speed is n c = 30 / R. In practice, the speed rate of the SAG mill is generally 65% to 80%.
Therefore, in the experiment, the speed was set to vary in 50% and 90% of the critical speed ( rad/s) for the crossover test as shown in Table 2.
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where g is m/s 2, R radius of the cylinder (m), r radius of the ball (m), and n c critical speed (rps). The operating speed of the ball mill is kept at 6580% of the critical speed. The lower
values are kept for the wet grinding in viscous solution, while a higher value is kept for dry grinding. Burr Mill or Plate Mill
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The two speeds being studied are 68% and 73% of critical speed in (′ diameter inside shell 16′ inside liners) ball mills. This study was over a period of four months. Grindability tests were run
on monthly composite samples of the feed to each mill. ... Ball mill size x (″ x 16′ diameter inside shell ′)
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Ball mills are predominantly used machines for grinding in the cement industry. Although ball mills have been used for more than one hundred years, the design is still being improved in order to
reduce the grinding ... Critical speed 76 % GM in II chamber 217 t Separator Sepax 450M222 Separator Cyclone 4 nos. Separator motor 300 kW ...
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Unknown Author - Puzzle Prime
If White is to play, can he always mate Black in 2 moves, regardless of the moves played before?
First, we notice that since Black made the last move, either the king or the rook has been moved, consequently rendering Black unable to castle. Now White plays Qa1 and no matter what is Black’s next
move, Qh8 gives check-mate.
Ping Pong Ball
Your last ping pong ball falls down into a narrow pipe embedded in concrete one foot deep. How can you get it out undamaged if all you have is your tennis paddle, your shoelaces, keys, wallet, and a
plastic water bottle, which does not fit into the pipe?
Using the plastic bottle, pour water into the pipe so that the ball will rise up.
Four grasshoppers start at the ends of a square in the plane. Every second one of them jumps over another one and lands on its other side at the same distance. Can the grasshoppers after finitely
many jumps end up at the vertices of a bigger square?
The answer is NO. In order to show this, assume they can and consider their reverse movement. Now the grasshoppers start at the vertices of some square, say with unit length sides, and end up at the
vertices of a smaller square. Create a lattice in the plane using the starting unit square. It is easy to see that the grasshoppers at all times will land on vertices of this lattice. However, it is
easy to see that every square with vertices coinciding with the lattice’s vertices has sides of length at least one. Therefore the assumption is wrong.
Fathers, Sons, and Fish
Two fathers and two sons went out fishing. Each of them catches two fish. However, they brought home only six fish. How so?
They were a son, his father, and his grandfather – 3 people in total.
Why do mirrors flip left and right but do not flip up and down?
Solution coming soon.
No Body, No Nose
What do you call a person with no body and no nose?
The answer is NOBODY KNOWS (no-body-nose).
The Connect Game
Two friends are playing the following game:
They start with 10 nodes on a sheet of paper and, taking turns, connect any two of them which are not already connected with an edge. The first player to make the resulting graph connected loses.
Who will win the game?
Remark: A graph is “connected” if there is a path between any two of its nodes.
The first player has a winning strategy.
His strategy is with each turn to keep the graph connected, until a single connected component of 6 or 7 nodes is reached. Then, his goal is to make sure the graph ends up with either connected
components of 8 and 2 nodes (8-2 split), or connected components of 6 and 4 nodes (6-4 split). In both cases, the two players will have to keep connecting nodes within these components, until one of
them is forced to make the graph connected. Since the number of edges in the components is either C^8_2+C^2_2=29, or C^6_2+C^4_2=21, which are both odd numbers, Player 1 will be the winner.
Once a single connected component of 6 or 7 nodes is reached, there are multiple possibilities:
1. The connected component has 7 nodes and Player 2 connects it to one of the three remaining nodes. Then, Player 1 should connect the remaining two nodes with each other and get an 8-2 split.
2. The connected component has 7 nodes and Player 2 connects two of the three remaining nodes with each other. Then, Player 1 should connect the large connected component to the last remaining node
and get an 8-2 split.
3. The connected component has 7 nodes and Player 2 makes a connection within it. Then, Player 1 also must connect two nodes within the component. Since the number of edges in a complete graph with
seven nodes is C^7_2=21, eventually Player 2 will be forced to make a move of type 1 or 2.
4. The connected component has 6 nodes and Player 2 connects it to one of the four remaining nodes. Then, Player 1 should make a connection within the connected seven nodes and reduce the game to
cases 1 to 3 above.
5. The connected component has 6 nodes and Player 2 connects two of the four remaining nodes. Then, Player 1 should connect the two remaining nodes with each other. The game is reduced to a 6-2-2
split which eventually will turn into either an 8-2 split, or a 6-4 split. In both cases Player 1 will win, as explained above.
Seven Letters Sequence
Find a seven-letter sequence to fill in each of the three empty spaces and form a meaningful sentence.
The ★★★★★★★ surgeon was ★★★ ★★★★ to operate, because there was ★★ ★★★★★.
The sequence is NOTABLE:
The NOTABLE surgeon was NOT ABLE to operate, because there was NO TABLE.
Glow and Shine
There is a property that applies to all words in the first list and to none in the words in the second list. What is it?
• GLOW, ALMOST, BIOPSY, GHOST, EMPTY, BEGIN
• SHINE, BARELY, VIVISECTION, APPARITION, VACANT, START
The words in the first list are called “Abecederian”, i.e. their letters are in alphabetical order.
Escaping the Kingdom
A long time ago there was a kingdom, isolated from the world. There was only one way to and from the kingdom, namely through a long bridge. The king ordered the execution of anyone caught fleeing the
kingdom on the bridge and the banishment of anyone caught sneaking into the kingdom.
The bridge was guarded by one person, who was taking a 10-minute break inside his cabin every round hour. Fifteen minutes were needed for a person to cross the bridge and yet, one woman managed to
escape the kingdom. How did she do it?
Once the guard entered the cabin, the woman started crossing the bridge for 9 minutes, and then turned around and pretended to be going in the opposite direction for one more minute. When the guard
caught her, she said she was trying to enter the kingdom, so he banished her away. | {"url":"https://www.puzzleprime.com/author/unknown-author/page/2/","timestamp":"2024-11-08T11:30:27Z","content_type":"text/html","content_length":"195819","record_id":"<urn:uuid:7243247f-942c-4f48-bce5-aa1ceddcc15a>","cc-path":"CC-MAIN-2024-46/segments/1730477028059.90/warc/CC-MAIN-20241108101914-20241108131914-00250.warc.gz"} |
How to check hypotheses with bootstrap and Apache Spark? - DataScienceCentral.com
Guest blog post by Dmitry Petrov. Originally posted here.
There is a feature I really like in Apache Spark. Spark can process data out of memory in my local machine even without a cluster. Good news for those who process data sets bigger than the memory
size that currently have. From time to time, I have this issue when I work with hypothesis testing.
For hypothesis testing I usually use statistical bootstrapping techniques. This method does not require any statistical knowledge and is very easy to understand. Also, this method is very simple to
implement. There are no normal distributions and student distributions from your statistical courses, only some basic coding skills. Good news for those who doesn’t like statistics. Spark and
bootstrapping is a very powerful combination which can help you check hypotheses in a large scale.
1. Bootstrap methods
The most common application with bootstrapping is calculating confidence intervals and you can use these confidence intervals as a part of the hypotheses checking process. There is a very simple idea
behind bootstrapping – sample your data set size N for hundreds or even thousands times with the replacement (this is important) and calculate the estimated metrics for each of the hundreds\thousands
subset. This process gives you a histogram which is an actual distribution for your data. Then, you can use this actual distribution for hypothesis testing.
The beauty of this method is the actual distribution histogram. In a classical statistical approach, you need to approximate a distribution of your data by normal distribution and calculate z-scores
or student-scores based on theoretical distributions. With the actual distribution from the first step it is easy to calculate 2.5% percentile and 97.5% percentiles and this would be your actual
confidence interval. That’s it! Confident interval with almost no math.
2. Choosing the right hypothesis
Choosing right hypotheses is only the tricky part in this analytical process. This is a question you ask the data and you cannot automate that. Hypotheses testing is a part of the analytical process
and isn’t usual for machine learning experts. In machine learning you ask an algorithm to build a model\structure which is sometimes called hypothesis and you are looking for the best hypotheses
which correlates your data and labels.
In the analytics process, knowing the correlation is not enough, you should know the hypothesis from the get-go and the question is – if the hypothesis is correct and what is your level of
If you have a correct hypotheses it is easy to check the hypotheses based on the bootstrapping approach. For example let’s try to check the hypothesis in which we take an average for some feature in
your dataset that is equal to 30.0. We should start with a null hypothesis H0 which we try to reject and an alternative hypothesis H1:
H0: mean(A) == 30.0
H1: meanA() != 30.0
If we fail to reject H0 we will take this hypothesis as ground truth. That’s what we need. If we don’t – then we should come up with a better hypothesis (mean(A) == 40).
3. Checking hypotheses
For the hypotheses checking we can simply calculate the confidence interval for dataset A by sampling and calculating 95% confidence interval. If the interval does not contain 30.0 then your
hypotheses H0 was rejected.
Obviously, this confident interval starts with 2.5% and ends 97.5% which gives us 95% of the items between this interval. In the sorted array of our observations we should find 2.5% and 97.5%
percentiles: p1 and p2. If p1 <= 30.0 <= p2, then we weren’t able to reject H0. So, we can suppose that H0 is the truth.
4. Apache Spark code
Implementation of bootstrapping in this particular case is straight forward.
import scala.util.Sorting.quickSort
def getConfInterval(input: org.apache.spark.rdd.RDD[Double], N: Int, left: Double, right:Double)
: (Double, Double) = {
// Simulate by sampling and calculating averages for each of subsamples
val hist = Array.fill(N){0.0}
for (i <- 0 to N-1) {
hist(i) = input.sample(withReplacement = true, fraction = 1.0).mean
// Sort the averages and calculate quantiles
val left_quantile = hist((N*left).toInt)
val right_quantile = hist((N*right).toInt)
return (left_quantile, right_quantile)
Because I did not find any good open datasets for the large scale hypotheses testing problem, let’s use skewdata.csv dataset from the book “Statistics: An Introduction Using R”. You can find this
dataset in this archive. It is not perfect but will work in a pinch.
val dataWithHeader = sc.textFile(“zipped/skewdata.csv”)
val header = dataWithHeader.first
val data = dataWithHeader.filter( _ != header ).map( _.toDouble )
val (left_qt, right_qt) = getConfInterval(data, 1000, 0.025, 0.975)
val H0_mean = 30
if (left_qt < H0_mean && H0_mean < right_qt) {
println(“We failed to reject H0. It seems like H0 is correct.”)
} else {
println(“We rejected H0”)
We have to understand the difference between “filed to reject H0” and “proof H0”. A failing to reject a hypothesis gives you a pretty strong level of evidence that the hypothesis is correct and you
can use this information in your decision making process but this is not an actual proof.
5. Equal means code example
Another type of hypotheses – check if the means of the two datasets are different. This leads us to the usual design of experiment questions – if you apply some change in your web system (user
interface change for example) would your click rate change in a positive direction?
Let’s create a hypothesis:
H0: mean(A) == mean(B)
H1: mean(A) > mean(B)
It is not easy to find H1 for this hypothesis which we can prove. Let’s change this hypothesis around a little bit:
Ho’: mean(A-B) == 0
H1: mean(A-B) > 0
Now we can try to reject H0′.
def getConfIntervalTwoMeans(input1: org.apache.spark.rdd.RDD[Double],
input2: org.apache.spark.rdd.RDD[Double],
N: Int, left: Double, right:Double)
: (Double, Double) = {
// Simulate average of differences
val hist = Array.fill(N){0.0}
for (i <- 0 to N-1) {
val mean1 = input1.sample(withReplacement = true, fraction = 1.0).mean
val mean2 = input2.sample(withReplacement = true, fraction = 1.0).mean
hist(i) = mean2 – mean1
// Sort the averages and calculate quantiles
val left_quantile = hist((N*left).toInt)
val right_quantile = hist((N*right).toInt)
return (left_quantile, right_quantile)
We should change 2.5% and 97.5% percentiles in the interval to 5% percentile in the left side only because of one-side (one-tailed) hypothesis testing. And an actual code as an example:
// Let’s try to check the same dataset with itself. Ha-ha.
val (left_qt, right_qt) = getConfIntervalTwoMeans(data, data, 1000, 0.05, 0.95)
// A condition was changed because of one-tailed test.
if (left_qt > 0) {
println(“We failed to reject H0. It seems like H0 is correct.”
} else {
println(“We rejected H0”)
Bootstrapping methods are very simple for understanding and implementation. They are intuitively simple and you don’t need any deep knowledge of statistics. Apache Spark can help you implement these
methods in a large scale.
As I mentioned previously it is not easy to find a good open large dataset for hypotheses testing. Please share with our community if you have one or come across one.
My code is shared in this Scala file.
About the Author:
Dmitry Petrov is a Data Scientist at Microsoft (Bellevue, Washington), working with the BingAds, Relevance and Revenue team. He earned his PhD in computer science from Saint Petersburg State
Electrotechnical University. He is also a scientific author, who received awards and developed a patent. | {"url":"https://www.datasciencecentral.com/how-to-check-hypotheses-with-bootstrap-and-apache-spark/","timestamp":"2024-11-04T02:18:38Z","content_type":"text/html","content_length":"162078","record_id":"<urn:uuid:96d7263f-08f8-4282-863b-a2311d66bd88>","cc-path":"CC-MAIN-2024-46/segments/1730477027809.13/warc/CC-MAIN-20241104003052-20241104033052-00078.warc.gz"} |
Run this notebook online: or Colab:
11.7. Adagrad¶
Let us begin by considering learning problems with features that occur infrequently.
11.7.1. Sparse Features and Learning Rates¶
Imagine that we are training a language model. To get good accuracy we typically want to decrease the learning rate as we keep on training, usually at a rate of \(\mathcal{O}(t^{-\frac{1}{2}})\) or
slower. Now consider a model training on sparse features, i.e., features that occur only infrequently. This is common for natural language, e.g., it is a lot less likely that we will see the word
preconditioning than learning. However, it is also common in other areas such as computational advertising and personalized collaborative filtering. After all, there are many things that are of
interest only for a small number of people.
Parameters associated with infrequent features only receive meaningful updates whenever these features occur. Given a decreasing learning rate we might end up in a situation where the parameters for
common features converge rather quickly to their optimal values, whereas for infrequent features we are still short of observing them sufficiently frequently before their optimal values can be
determined. In other words, the learning rate either decreases too quickly for frequent features or too slowly for infrequent ones.
A possible hack to redress this issue would be to count the number of times we see a particular feature and to use this as a clock for adjusting learning rates. That is, rather than choosing a
learning rate of the form \(\eta = \frac{\eta_0}{\sqrt{t + c}}\) we could use \(\eta_i = \frac{\eta_0}{\sqrt{s(i, t) + c}}\). Here \(s(i, t)\) counts the number of nonzeros for feature \(i\) that we
have observed up to time \(t\). This is actually quite easy to implement at no meaningful overhead. However, it fails whenever we do not quite have sparsity but rather just data where the gradients
are often very small and only rarely large. After all, it is unclear where one would draw the line between something that qualifies as an observed feature or not.
Adagrad by [Duchi et al., 2011] addresses this by replacing the rather crude counter \(s(i, t)\) by an aggregate of the squares of previously observed gradients. In particular, it uses \(s(i, t+1) =
s(i, t) + \left(\partial_i f(\mathbf{x})\right)^2\) as a means to adjust the learning rate. This has two benefits: first, we no longer need to decide just when a gradient is large enough. Second, it
scales automatically with the magnitude of the gradients. Coordinates that routinely correspond to large gradients are scaled down significantly, whereas others with small gradients receive a much
more gentle treatment. In practice this leads to a very effective optimization procedure for computational advertising and related problems. But this hides some of the additional benefits inherent in
Adagrad that are best understood in the context of preconditioning.
11.7.2. Preconditioning¶
Convex optimization problems are good for analyzing the characteristics of algorithms. After all, for most nonconvex problems it is difficult to derive meaningful theoretical guarantees, but
intuition and insight often carry over. Let us look at the problem of minimizing \(f(\mathbf{x}) = \frac{1}{2} \mathbf{x}^\top \mathbf{Q} \mathbf{x} + \mathbf{c}^\top \mathbf{x} + b\).
As we saw in Section 11.6, it is possible to rewrite this problem in terms of its eigendecomposition \(\mathbf{Q} = \mathbf{U}^\top \boldsymbol{\Lambda} \mathbf{U}\) to arrive at a much simplified
problem where each coordinate can be solved individually:
\[f(\mathbf{x}) = \bar{f}(\bar{\mathbf{x}}) = \frac{1}{2} \bar{\mathbf{x}}^\top \boldsymbol{\Lambda} \bar{\mathbf{x}} + \bar{\mathbf{c}}^\top \bar{\mathbf{x}} + b.\]
Here we used \(\mathbf{x} = \mathbf{U} \mathbf{x}\) and consequently \(\mathbf{c} = \mathbf{U} \mathbf{c}\). The modified problem has as its minimizer \(\bar{\mathbf{x}} = -\boldsymbol{\Lambda}^{-1}
\bar{\mathbf{c}}\) and minimum value \(-\frac{1}{2} \bar{\mathbf{c}}^\top \boldsymbol{\Lambda}^{-1} \bar{\mathbf{c}} + b\). This is much easier to compute since \(\boldsymbol{\Lambda}\) is a diagonal
matrix containing the eigenvalues of \(\mathbf{Q}\).
If we perturb \(\mathbf{c}\) slightly we would hope to find only slight changes in the minimizer of \(f\). Unfortunately this is not the case. While slight changes in \(\mathbf{c}\) lead to equally
slight changes in \(\bar{\mathbf{c}}\), this is not the case for the minimizer of \(f\) (and of \(\bar{f}\) respectively). Whenever the eigenvalues \(\boldsymbol{\Lambda}_i\) are large we will see
only small changes in \(\bar{x}_i\) and in the minimum of \(\bar{f}\). Conversely, for small \(\boldsymbol{\Lambda}_i\) changes in \(\bar{x}_i\) can be dramatic. The ratio between the largest and the
smallest eigenvalue is called the condition number of an optimization problem.
\[\kappa = \frac{\boldsymbol{\Lambda}_1}{\boldsymbol{\Lambda}_d}.\]
If the condition number \(\kappa\) is large, it is difficult to solve the optimization problem accurately. We need to ensure that we are careful in getting a large dynamic range of values right. Our
analysis leads to an obvious, albeit somewhat naive question: couldn’t we simply “fix” the problem by distorting the space such that all eigenvalues are \(1\). In theory this is quite easy: we only
need the eigenvalues and eigenvectors of \(\mathbf{Q}\) to rescale the problem from \(\mathbf{x}\) to one in \(\mathbf{z} := \boldsymbol{\Lambda}^{\frac{1}{2}} \mathbf{U} \mathbf{x}\). In the new
coordinate system \(\mathbf{x}^\top \mathbf{Q} \mathbf{x}\) could be simplified to \(\|\mathbf{z}\|^2\). Alas, this is a rather impractical suggestion. Computing eigenvalues and eigenvectors is in
general much more expensive than solving the actual problem.
While computing eigenvalues exactly might be expensive, guessing them and computing them even somewhat approximately may already be a lot better than not doing anything at all. In particular, we
could use the diagonal entries of \(\mathbf{Q}\) and rescale it accordingly. This is much cheaper than computing eigenvalues.
\[\tilde{\mathbf{Q}} = \mathrm{diag}^{-\frac{1}{2}}(\mathbf{Q}) \mathbf{Q} \mathrm{diag}^{-\frac{1}{2}}(\mathbf{Q}).\]
In this case we have \(\tilde{\mathbf{Q}}_{ij} = \mathbf{Q}_{ij} / \sqrt{\mathbf{Q}_{ii} \mathbf{Q}_{jj}}\) and specifically \(\tilde{\mathbf{Q}}_{ii} = 1\) for all \(i\). In most cases this
simplifies the condition number considerably. For instance, the cases we discussed previously, this would entirely eliminate the problem at hand since the problem is axis aligned.
Unfortunately we face yet another problem: in deep learning we typically do not even have access to the second derivative of the objective function: for \(\mathbf{x} \in \mathbb{R}^d\) the second
derivative even on a minibatch may require \(\mathcal{O}(d^2)\) space and work to compute, thus making it practically infeasible. The ingenious idea of Adagrad is to use a proxy for that elusive
diagonal of the Hessian that is both relatively cheap to compute and effective—the magnitude of the gradient itself.
In order to see why this works, let us look at \(\bar{f}(\bar{\mathbf{x}})\). We have that
\[\partial_{\bar{\mathbf{x}}} \bar{f}(\bar{\mathbf{x}}) = \boldsymbol{\Lambda} \bar{\mathbf{x}} + \bar{\mathbf{c}} = \boldsymbol{\Lambda} \left(\bar{\mathbf{x}} - \bar{\mathbf{x}}_0\right),\]
where \(\bar{\mathbf{x}}_0\) is the minimizer of \(\bar{f}\). Hence the magnitude of the gradient depends both on \(\boldsymbol{\Lambda}\) and the distance from optimality. If \(\bar{\mathbf{x}} - \
bar{\mathbf{x}}_0\) didn’t change, this would be all that’s needed. After all, in this case the magnitude of the gradient \(\partial_{\bar{\mathbf{x}}} \bar{f}(\bar{\mathbf{x}})\) suffices. Since
AdaGrad is a stochastic gradient descent algorithm, we will see gradients with nonzero variance even at optimality. As a result we can safely use the variance of the gradients as a cheap proxy for
the scale of the Hessian. A thorough analysis is beyond the scope of this section (it would be several pages). We refer the reader to [Duchi et al., 2011] for details.
11.7.3. The Algorithm¶
Let us formalize the discussion from above. We use the variable \(\mathbf{s}_t\) to accumulate past gradient variance as follows.
\[\begin{split}\begin{aligned} \mathbf{g}_t & = \partial_{\mathbf{w}} l(y_t, f(\mathbf{x}_t, \mathbf{w})), \\ \mathbf{s}_t & = \mathbf{s}_{t-1} + \mathbf{g}_t^2, \\ \mathbf{w}_t & = \mathbf{w}_{t-1}
- \frac{\eta}{\sqrt{\mathbf{s}_t + \epsilon}} \cdot \mathbf{g}_t. \end{aligned}\end{split}\]
Here the operation are applied coordinate wise. That is, \(\mathbf{v}^2\) has entries \(v_i^2\). Likewise \(\frac{1}{\sqrt{v}}\) has entries \(\frac{1}{\sqrt{v_i}}\) and \(\mathbf{u} \cdot \mathbf{v}
\) has entries \(u_i v_i\). As before \(\eta\) is the learning rate and \(\epsilon\) is an additive constant that ensures that we do not divide by \(0\). Last, we initialize \(\mathbf{s}_0 = \mathbf
Just like in the case of momentum we need to keep track of an auxiliary variable, in this case to allow for an individual learning rate per coordinate. This does not increase the cost of Adagrad
significantly relative to SGD, simply since the main cost is typically to compute \(l(y_t, f(\mathbf{x}_t, \mathbf{w}))\) and its derivative.
Note that accumulating squared gradients in \(\mathbf{s}_t\) means that \(\mathbf{s}_t\) grows essentially at linear rate (somewhat slower than linearly in practice, since the gradients initially
diminish). This leads to an \(\mathcal{O}(t^{-\frac{1}{2}})\) learning rate, albeit adjusted on a per coordinate basis. For convex problems this is perfectly adequate. In deep learning, though, we
might want to decrease the learning rate rather more slowly. This led to a number of Adagrad variants that we will discuss in the subsequent chapters. For now let us see how it behaves in a quadratic
convex problem. We use the same problem as before:
\[f(\mathbf{x}) = 0.1 x_1^2 + 2 x_2^2.\]
We are going to implement Adagrad using the same learning rate previously, i.e., \(\eta = 0.4\). As we can see, the iterative trajectory of the independent variable is smoother. However, due to the
cumulative effect of \(\boldsymbol{s}_t\), the learning rate continuously decays, so the independent variable does not move as much during later stages of iteration.
%load ../utils/djl-imports
%load ../utils/plot-utils
%load ../utils/Functions.java
%load ../utils/GradDescUtils.java
%load ../utils/Accumulator.java
%load ../utils/StopWatch.java
%load ../utils/Training.java
%load ../utils/TrainingChapter11.java
float eta = 0.4f;
Function<Float[], Float[]> adagrad2d = (state) -> {
Float x1 = state[0], x2 = state[1], s1 = state[2], s2 = state[3];
float eps = (float) 1e-6;
float g1 = 0.2f * x1;
float g2 = 4 * x2;
s1 += g1 * g1;
s2 += g2 * g2;
x1 -= eta / (float) Math.sqrt(s1 + eps) * g1;
x2 -= eta / (float) Math.sqrt(s2 + eps) * g2;
return new Float[]{x1, x2, s1, s2};
BiFunction<Float, Float, Float> f2d = (x1, x2) -> 0.1f * x1 * x1 + 2 * x2 * x2;
GradDescUtils.showTrace2d(f2d, GradDescUtils.train2d(adagrad2d, 20));
Tablesaw not supporting for contour and meshgrids, will update soon
As we increase the learning rate to \(2\) we see much better behavior. This already indicates that the decrease in learning rate might be rather aggressive, even in the noise-free case and we need to
ensure that parameters converge appropriately.
eta = 2;
GradDescUtils.showTrace2d(f2d, GradDescUtils.train2d(adagrad2d, 20));
Tablesaw not supporting for contour and meshgrids, will update soon
11.7.4. Implementation from Scratch¶
Just like the momentum method, Adagrad needs to maintain a state variable of the same shape as the parameters.
NDList initAdagradStates(int featureDimension) {
NDManager manager = NDManager.newBaseManager();
NDArray sW = manager.zeros(new Shape(featureDimension, 1));
NDArray sB = manager.zeros(new Shape(1));
return new NDList(sW, sB);
public class Optimization {
public static void adagrad(NDList params, NDList states, Map<String, Float> hyperparams) {
float eps = (float) 1e-6;
for (int i = 0; i < params.size(); i++) {
NDArray param = params.get(i);
NDArray state = states.get(i);
// Update param
Compared to the experiment in Section 11.5 we use a larger learning rate to train the model.
AirfoilRandomAccess airfoil = TrainingChapter11.getDataCh11(10, 1500);
public TrainingChapter11.LossTime trainAdagrad(float lr, int numEpochs) throws IOException, TranslateException {
int featureDimension = airfoil.getColumnNames().size();
Map<String, Float> hyperparams = new HashMap<>();
hyperparams.put("lr", lr);
return TrainingChapter11.trainCh11(Optimization::adagrad,
hyperparams, airfoil, featureDimension, numEpochs);
TrainingChapter11.LossTime lossTime = trainAdagrad(0.1f, 2);
loss: 0.243, 0.083 sec/epoch
11.7.5. Concise Implementation¶
We can use the Adagrad algorithm in DJL by creating an instance of Adagrad from Optimizer. Then we can pass it into our trainConciseCh11() function defined in chapter 11.5 to train with it!
Tracker lrt = Tracker.fixed(0.1f);
Optimizer adagrad = Optimizer.adagrad().optLearningRateTracker(lrt).build();
TrainingChapter11.trainConciseCh11(adagrad, airfoil, 2);
INFO Training on: 1 GPUs.
INFO Load MXNet Engine Version 1.9.0 in 0.064 ms.
Training: 100% |████████████████████████████████████████| Accuracy: 1.00, L2Loss: 0.26
loss: 0.243, 0.169 sec/epoch
11.7.6. Summary¶
• Adagrad decreases the learning rate dynamically on a per-coordinate basis.
• It uses the magnitude of the gradient as a means of adjusting how quickly progress is achieved - coordinates with large gradients are compensated with a smaller learning rate.
• Computing the exact second derivative is typically infeasible in deep learning problems due to memory and computational constraints. The gradient can be a useful proxy.
• If the optimization problem has a rather uneven uneven structure Adagrad can help mitigate the distortion.
• Adagrad is particularly effective for sparse features where the learning rate needs to decrease more slowly for infrequently occurring terms.
• On deep learning problems Adagrad can sometimes be too aggressive in reducing learning rates. We will discuss strategies for mitigating this in the context of Section 11.10.
11.7.7. Exercises¶
1. Prove that for an orthogonal matrix \(\mathbf{U}\) and a vector \(\mathbf{c}\) the following holds: \(\|\mathbf{c} - \mathbf{\delta}\|_2 = \|\mathbf{U} \mathbf{c} - \mathbf{U} \mathbf{\delta}\|_2
\). Why does this mean that the magnitude of perturbations does not change after an orthogonal change of variables?
2. Try out Adagrad for \(f(\mathbf{x}) = 0.1 x_1^2 + 2 x_2^2\) and also for the objective function was rotated by 45 degrees, i.e., \(f(\mathbf{x}) = 0.1 (x_1 + x_2)^2 + 2 (x_1 - x_2)^2\). Does it
behave differently?
3. Prove Gerschgorin’s circle theorem which states that eigenvalues \(\lambda_i\) of a matrix \(\mathbf{M}\) satisfy \(|\lambda_i - \mathbf{M}_{jj}| \leq \sum_{k \neq j} |\mathbf{M}_{jk}|\) for at
least one choice of \(j\).
4. What does Gerschgorin’s theorem tell us about the eigenvalues of the diagonally preconditioned matrix \(\mathrm{diag}^{-\frac{1}{2}}(\mathbf{M}) \mathbf{M} \mathrm{diag}^{-\frac{1}{2}}(\mathbf
5. Try out Adagrad for a proper deep network, such as Section 6.6 when applied to Fashion MNIST.
6. How would you need to modify Adagrad to achieve a less aggressive decay in learning rate? | {"url":"https://d2l.djl.ai/chapter_optimization/adagrad.html","timestamp":"2024-11-07T03:07:41Z","content_type":"application/xhtml+xml","content_length":"83481","record_id":"<urn:uuid:2b48b95c-8a55-4b8f-a6f1-8a4e03195c19>","cc-path":"CC-MAIN-2024-46/segments/1730477027951.86/warc/CC-MAIN-20241107021136-20241107051136-00642.warc.gz"} |
Kimwoods Can You Post The Work As Discussed - topchoicewriters.com
Complete both Part A and Part B below.
Part A
Before completing the following questions, be sure to have read Appendix C and the Statistical Software Resources at the ends of Chapters 2 and 3 from Statistics Plain and Simple.
Highlight the required answers to the question in your Excel output.
1. Using Microsoft® Excel®, enter the following data from the 40 participants by first creating a variable labeled “Score”. Next, compute the mean, median, and mode for the following set of 40
reading scores:
2. Imagine you are the assistant manager of a fast food store. Part of your job is to report which special is selling best to the store manager at the end of each day. Use your knowledge of
descriptive statistics and write one paragraph to let the store manager know what happened today. Use the following data.
Special number Sold Cost
Huge Burger 20 $2.95
Baby Burger 18 $1.49
Chicken Littles 25 $3.50
Porker Burger 19 $2.95
Yummy Burger 17 $1.99
Coney Dog 20 $1.99
Total specials sold 119
3. Suppose you are working with a data set that has some different (much larger or much smaller than the rest of the data) scores. What measure of central tendency (mean, median or mode) would you
use and why?
4. During the course of a semester, 10 students in Mr. Smith’s class took three exams. Use Microsoft® Excel® to compute all the descriptive statistics for the following set of three test scores over
the course of a semester. Which test had the highest average score? Which test had the smallest amount of variability? How would you interpret the differences between exams, and note the range,
means, and standard deviations over time?
5. For each of the following, indicate whether you would use a pie, line, or bar chart, and why:
a. The proportion of freshmen, sophomores, juniors, and seniors in a particular university
b. Change in GPA over four semesters
c. Number of applicants for four different jobs
d. Reaction time to different stimuli
e. Number of scores in each of 10 categories
6. Using the data from question 1, create a frequency table and a histogram in Microsoft® Excel®.
Part B
Answer the questions below. Be specific and provide examples when relevant.
Cite any sources consistent with APA guidelines.
Question Answer
What are statistics and how are they used in the behavioral sciences? Your answer should be 100 to 175 words.
Providing examples of each, compare and contrast the four levels of measurement. Your answer should be 175 to 350 words.
Differentiate between descriptive and inferential statistics. What information do they provide? What are their similarities and differences? Your answer should be 175 to 350 words.
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2021-02-07 09:53:502021-02-07 09:53:50Kimwoods Can You Post The Work As Discussed | {"url":"https://topchoicewriters.com/kimwoods-can-you-post-the-work-as-discussed/","timestamp":"2024-11-04T17:02:23Z","content_type":"text/html","content_length":"50615","record_id":"<urn:uuid:4fbbd2a4-11ea-400d-a9ce-5125499c36ee>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00517.warc.gz"} |
Calculate Board Feet and Optional Cost
┃ This converter requires the use of Javascript enabled and capable browsers. This calculator is to determine the number of board feet for a given size of board, based on the formula (((Depth in ┃
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┃ both and one MUST be changed. Leading zeros do not effect calculations. Each board must have a width in inches and a depth, also in inches. For instance, a 2 x 4 is (by definition only) 2 inches ┃
┃ in depth and 4 inches in width. Those are the defaults for each since 2 x 4s are the most common lumber. It can be any length. For instance, 5 feet can be entered as 5 in the feet field or 60 in ┃
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┃ boards, you would change that to 5. If they were all 8 foot studs, you would fill in the length as 8 feet, 0 inches, 2 depth, and 4 width. You may also use the quantity field to help indicate the ┃
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┃ foot 2 x 4s, you would enter 50 feet, 0 inches, 2 depth, 4 width and a quantity of 1. The total number of lineal feet will be displayed as one of the results. If you wish to know the total cost ┃
┃ of the boardfootage, and you know the cost per boardfoot, enter that value in the price field (as for example 1.25 for a dollar and a quarter per board foot) and a total cost will be calculated ┃
┃ in dollars and cents. The Reset button assigns default values to all fields and clears the totals. A value of NaN in either of the totals fields is an error message and indicates an improper ┃
┃ entry in one of the factors. To calculate and total quantities of specific sizes of boards, use our Boardfeet Accumulator. To convert a number of boardfeet to lineal feet, use our Boardfoot To ┃
┃ Lineal Foot Converter. ┃
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┃ Version 2.2.0 ┃ | {"url":"http://www.csgnetwork.com/boardftcalc.html","timestamp":"2024-11-05T12:56:45Z","content_type":"text/html","content_length":"17698","record_id":"<urn:uuid:f62e0b7a-a58c-4913-bd6f-c66acf763b1a>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00681.warc.gz"} |
Video - Sheaf quantization of Hamiltonian isotopies and non-displacability problems
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2. The lace expansion in the past, present and future
3. Fano manifolds of Calabi-Yau Type
4. Sums of squares in quadratic number rings
5. Entropies on covers of compact manifolds
6. Quantum Dynamics in the Mean-Field and Semiclassical Regime
7. Random walks in spaces of negative curvature
8. The Shape of Data
9. The significance of dimensions in mathematics
10. Topological aspects in the theory of aperiodic solids and tiling spaces
11. Noncommutative Surfaces
12. Conformal field theory and noncommutative geometry
13. Faithful representations of Chevalley groups over quotient rings of non-Archimedean local fields
14. Analytic torsion and mirror symmetry
15. Deformation spaces of Kleinian groups and beyond
16. A-infinity functor and topological field theory
17. Number theoretic results in a family
18. Quasi-homomorphisms into non-commutative groups
19. Conservation laws and differential geometry
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samplesizeCMH: Sample Size Calculation for the Cochran-Mantel-Haenszel Test
by Paul W. Egeler M.S.
This package provides functions relating to power and sample size calculation for the CMH test. There are also several helper functions for interconverting probability, odds, relative risk, and odds
ratio values.
Please see the package website for more information on how this package is used, including documentation and vignettes.
The Cochran Mantel Haenszel Test
The Cochran-Mantel-Haenszel test (CMH) is an inferential test for the association between two binary variables, while controlling for a third confounding nominal variable. Two variables of interest,
X and Y, are compared at each level of the confounder variable Z and the results are combined, creating a common odds ratio. Essentially, the CMH test examines the weighted association of X and Y.
The CMH test is a common technique in the field of biostatistics, where it is often used for case-control studies.
Sample Size Calculation
Given a target power which the researcher would like to achieve, a calculation can be performed in order to estimate the appropriate number of subjects for a study. The power.cmh.test function
calculates the required number of subjects per group to achieve a specified power for a Cochran-Mantel-Haenszel test.
Power Calculation
Researchers interested in estimating the probability of detecting a true positive result from an inferential test must perform a power calculation using a known sample size, effect size, significance
level, et cetera. The power.cmh.test function can compute the power of a CMH test, given parameters from the experiment.
Installation of the CRAN release can be done with install.packages(). From the R console:
Downloading and installing the latest version from GitHub is facilitated by remotes. To do so, type the following into your R console: | {"url":"https://cloud.r-project.org/web/packages/samplesizeCMH/readme/README.html","timestamp":"2024-11-06T00:48:28Z","content_type":"application/xhtml+xml","content_length":"7681","record_id":"<urn:uuid:5bce79b9-29ed-435a-a4ed-ca5dfed593b1>","cc-path":"CC-MAIN-2024-46/segments/1730477027906.34/warc/CC-MAIN-20241106003436-20241106033436-00129.warc.gz"} |
Видеотека: S. O. Gorchinskiy, Milnor $K$-groups of nilpotent extensions
Аннотация: The talk is based on a series of common works with Dimitrii Tyurin and with Denis Osipov. We prove a version of the famous Goodwillie's theorem with algebraic $K$-groups being replaced by
Milnor $K$-groups. Namely, given a commutative ring $R$ with a nilpotent ideal $I$, $I^N=0$, such that the quotient $R/I$ splits, we study relative Milnor $K$-groups $K^M_{n+1}(R,I)$, $n\geqslant 0$.
Provided that the ring $R$ has enough invertible elements in a sense, these groups are related to the quotient of the module of relative differential forms $\Omega^n_{R,I}/d\,\Omega^{n-1}_{R,I}$.
This holds in two different cases: when $N!$ is invertible in $R$ and when $R$ is a complete $p$-adic ring with a lift of Frobenius. However, the approaches and constructions are different in these
Язык доклада: английский | {"url":"https://m.mathnet.ru/php/presentation.phtml?option_lang=rus&presentid=35331","timestamp":"2024-11-13T18:05:27Z","content_type":"text/html","content_length":"8681","record_id":"<urn:uuid:b675d807-9178-4507-8986-d7b8fcff18ca>","cc-path":"CC-MAIN-2024-46/segments/1730477028387.69/warc/CC-MAIN-20241113171551-20241113201551-00018.warc.gz"} |
The Star Puzzle | R-bloggersThe Star Puzzle
[This article was first published on
Econometrics by Simulation
, and kindly contributed to
]. (You can report issue about the content on this page
) Want to share your content on R-bloggers?
if you have a blog, or
if you don't.
The Star Puzzle is a puzzle presented on
The Math Forum
. I became aware of this problem by noticing the article and solution posted on
Quantitative Decisions
article section. It asks the question, “How many triangles, quadrilaterals, and irregular hexagons can we form from a star of David?”
In this post I will solve these questions using R as well as solve the more basic question, “How to we find the geometric placement of each of our nodes so as to graph all of our shapes perfectly?”
First off, we note that the star is constructed of two intersecting equilateral triangles which when connecting each intersection points creates a total of 12 smaller equilateral triangles.
Let us first define the origin as (0,0) and define the length of one of the larger sides as 9a (but for future reference just 9). The a is there to indicate that we can arbitrarily scale the star at
any time.
Now we want to find all of the nodes (corner pieces in all of the 12 triangles) on the star. Some values we know already, for instance the x values are all -4.5, 0, or 4.5 because the star is
centered at the origin and each side is 9. Also, we know the length of each side of the smaller triangles since 3 smaller triangle lengths spans the larger triangle length thus length of each
smaller triangle is 3, giving us the coordinates of our first two nodes (-3,0) and (3,0).
Next we need calculate the height of one of the triangles. Using the Pythagorean Theorem.
Now we can fill in all of the values.
Let us send these values to R.
# First let us set the margins to be zero
par(mar = rep(0, 4))
# First we will come up with a node list
node <-
rbind(c(0,5.2), c(1.5,2.6), c(4.5,2.6), c(3,0),
c(4.5,-2.6), c(1.5,-2.6), c(0,-5.2), c(-1.5,-2.6),
c(-4.5,-2.6),c(-3,0), c(-4.5,2.6), c(-1.5,2.6),
plot(node, type="n", ylim=c(min(node[,2]), max(node[,2])+.1))
for (i in 1:13) text(node[i,1],node[i,2]+.3, i)
# First the perimeter
# Now the regular hexigon.
lines(node[c(seq(2,12,2),2),], lwd=2)
# Finally the lines connecting the intersection points.
lines(node[c(2,8),] , lwd=2)
lines(node[c(4,10),], lwd=2)
lines(node[c(6,12),], lwd=2)
# Now we are ready to try to solve the challenge. Let us first start by defining
# a list of connections for each node.
connections <- list(
c(12,2), # for 1
c(1,12,13,4,3), # for 2
c(2,4), # for 3
c(2,3,5,6,13), # for 4
c(4,6), # for 5
c(7,8,13,4,5), # for 6
c(8,6), # for 7
c(9,10,13,6,7), # for 8
c(8,10), # for 9
c(11,12,13,8,9), # for 10
c(10,12), # for 11
c(11,1,2,13,10), # for 12
c(2,4,6,8,10,12) # for 13
# In order to count the number of sides we will need to define a vector that
# collapses nodes on the same side.
collapser <- rbind(
# In the counting of sides. If there is a single node between
# any of the above sets
# we will remove that node. Thus 1,2,4 will become 1,4 which counts as one side.
# We only need to run the matching algorithm once because
# 1,2,4,5 -> drops 2 and 4 -> 1,5
# Let us create a function to count the number of sides.
nsides <- function(nodelist, collapser) {
# A list of nodes to keep
keeper <- rep(T,length(nodelist))
# Only evaluate sides if list of nodes >2
if (length(nodelist)>2) # Cycle through each set of 3 nodes
for (i in 1:(length(nodelist)))
for (ii in 1:nrow(collapser)) {
if (i+1 < length(nodelist)) {
if ((nodelist[i]==collapser[ii,1]&nodelist[i+2]==collapser[ii,2])|
(nodelist[i] ==collapser[ii,2]&nodelist[i+2]==collapser[ii,1]))
keeper[i+1] <- F
if (i+1 == length(nodelist))
if ((nodelist[i]==collapser[ii,1]&nodelist[1]==collapser[ii,2])|
(nodelist[i] ==collapser[ii,2]&nodelist[1]==collapser[ii,1]))
keeper[i+1] <- F
if (i == length(nodelist))
if ((nodelist[i]==collapser[ii,1]&nodelist[2]==collapser[ii,2])|
(nodelist[i] ==collapser[ii,2]&nodelist[2]==collapser[ii,1]))
keeper[1] <- F
# Return both the list of condensed nodes and the number of sides
list(node=nodelist[keeper], sides=sum(keeper))
# Create a simple function for plotting node shapes
plot.node <- function(s, nodes=node, ntext=T, lwd=2, clear=F,
col = rgb(0, 0, 0,0.15),border="black", ...) {
if (clear) plot(nodes[s,], type="n", xaxt='n', yaxt='n',
ylim=c(min(nodes[s,2]), max(nodes[s,2])+.2), ...)
polygon(nodes[s,1],nodes[s,2],lwd=2, col = col, border=border)
if (ntext) for (i in s) text(nodes[i,1],nodes[i,2]+.3, i)
nsides(c(1,2,4,13,10,12), collapser)
(s <- nsides(c(1,2,3,4,13,10,12), collapser))
plot.node(s[[1]], clear=T, border="white")
(s <- nsides(c(12,1,2,4,5,6,13,12), collapser))
(s <- nsides(c(1,2,13,8,9,10,12), collapser))
# Rotate matcher
rmatch <- function(first,second) {
for (i in 1:length(first)) {
mixer <- c(second[-(1:i)],second[1:i])
if (all(first==mixer)|all(rev(first)==mixer)) return(T)
F #
rmatch(1:5, c(2:5,1))
rmatch(5:1, c(2:5,1))
# We will use a recursive algorithm to search each node and each connection
# to each node.
nodesearch <- function(nodelist, connections, target.sides,
collapser, noisily=F) {
# Calculate number of side and shortest node list
fsides <- nsides(nodelist[-1], collapser)
# Calculate possible connections to the current node
cconct <- connections[[(rev(nodelist)[1])]]
# Calculate available connections to current node given
# some nodes have already been used.
aconct <- cconct[!(cconct %in% nodelist[-1])]
if ((fsides[[2]]>target.sides+1)|length(aconct)==0|
# If noisily option is selected then a print screen will occure any
# time there is a end of path.
if (noisily) {
print(paste("sides", fsides[[2]]))
if (nodelist[1]==rev(nodelist)[1]&fsides[[2]]==target.sides) {
# Search if the current solution is a duplicate of another
dup <- F
for (i in ret.ls) if (rmatch(fsides[[1]], i)) dup <- T
if (!dup) ret.ls[[length(ret.ls)+1]] <<- rev(fsides[[1]])
} else {
for (i in aconct) nodesearch(c(nodelist,i), connections, target.sides,
# Stores the number of polygons starting at node 3 with a specific number
# of sides. We know there are no 2 sided polygons.
nodecount <- c(sides=2, matches=0)
# This will return all possible three sided objects
# (triangles in this case) which pass through node 3
sides <- 3; ret.ls <- list()
nodesearch(3, connections, sides, collapser)
# Create a counter of number of polygons including point 3
(nodecount <- rbind(nodecount, c(sides, length(ret.ls))))
plot.node(1:12, col=gray(1), ntext=F, clear=T)
for (i in ret.ls) plot.node(i, ntext=T, clear=F)
# 3 exist
# Four sided
sides <- 4; ret.ls <- list()
nodesearch(3, connections, sides, collapser)
(nodecount <- rbind(nodecount, c(sides, length(ret.ls))))
plot.node(1:12, col=gray(1), ntext=F, clear=T)
for (i in ret.ls) plot.node(i, ntext=T, clear=F)
# 10 exist
# Five sided
sides <- 5; ret.ls <- list()
nodesearch(3, connections, sides, collapser)
(nodecount <- rbind(nodecount, c(sides, length(ret.ls))))
plot.node(1:12, col=gray(1), ntext=F, clear=T)
for (i in ret.ls[1:2]) plot.node(i, ntext=T, clear=F)
# 19 exist
# Just plotting two of the irregular pentagons
# Six Sided
sides <- 6; ret.ls <- list()
nodesearch(3, connections, sides, collapser)
(nodecount <- rbind(nodecount, c(sides, length(ret.ls))))
plot.node(1:12, col=gray(1), ntext=F, clear=T)
for (i in ret.ls[1:2]) plot.node(i, ntext=T, clear=F)
# 30 exist
# Plotting two of the irregular hexigons
for (sides in 7:16) {
ret.ls <- list()
nodesearch(3, connections, sides, collapser)
nodecount <- rbind(nodecount, c(sides, length(ret.ls)))
par(mar = c(4,4,4,1))
plot(nodecount, type="b",
main="Number of Polygons Peaks at 7\n(Starting at node 3)")
# Looking at all nodes
# Matrix for full node count
fnodecount <- c(sides=2, matches=0)
# We can collect all possible triangles:
sides <- 3; ret.ls <- list()
for (i in 1:13) nodesearch(i, connections, sides, collapser)
(fnodecount <- rbind(fnodecount, c(sides, length(ret.ls))))
# Which yeilds a list of 20 possible triangles
par(mfrow=c(4,5), mar=rep(0,4))
for (i in ret.ls) {
plot.node(1:12, col=gray(1), border=gray(.7), ntext=F, clear=T)
plot.node(i, ntext=F, clear=F)
# As for quadralaterals
sides <- 4; ret.ls <- list()
for (i in 1:13) nodesearch(i, connections, sides, collapser)
(fnodecount <- rbind(fnodecount, c(sides, length(ret.ls))))
# 57 Quadralaterals
par(mfrow=c(6,5), mar=rep(0,4))
for (i in ret.ls) {
plot.node(1:12, col=gray(1), border=gray(.7), ntext=F, clear=T)
plot.node(i, ntext=F, clear=F)
# pentagons
sides <- 5; ret.ls <- list()
for (i in 1:13) nodesearch(i, connections, sides, collapser)
(fnodecount <- rbind(fnodecount, c(sides, length(ret.ls))))
# 60 pentagons
# and hexigons
# pentagons
sides <- 6; ret.ls <- list()
for (i in 1:13) nodesearch(i, connections, sides, collapser)
(fnodecount <- rbind(fnodecount, c(sides, length(ret.ls))))
# 100 hexigons
par(mfrow=c(10,10), mar=rep(0,4))
for (i in ret.ls) {
plot.node(1:12, col=gray(1), border=gray(.7), ntext=F, clear=T)
plot.node(i, ntext=F, clear=F)
# Notice that node one is used in the first three rows then node
# two is used in all other rows as well up till the last seven
# polygons.
# and septigons and more for (sides in 7:16) { ret.ls <- list() for (i in 1:13) nodesearch(i, connections, sides, collapser) print(fnodecount <- rbind(fnodecount, c(sides, length(ret.ls)))) }
par(mfrow=c(10,10), mar=rep(0,4)) plot(fnodecount, type="b", main="Number of Polygons Peaks at 6") lines(nnodecount, type="b", col="darkblue")
# The blue lower line is the number of polygons that use node 3 as origin
# -----------------------------------
# Overall the code seems to work well. The approach is easily generalizable and
# could calculate polygons even if nodes or connections were added or removed.
par(mar = c(0,0,0,0))
plot(node, type="n", ylim=c(min(node[,2]), max(node[,2])+.1))
for (i in 1:13) text(node[i,1],node[i,2]+.3, i)
lines(node[c(seq(2,12,2),2),], lwd=2)
lines(node[c(2,8),] , lwd=2)
lines(node[c(4,10),], lwd=2)
lines(node[c(6,12),], lwd=2)
# Adding in the lines to the outer hexigon
lines(node[c(seq(1,11,2),1),], lwd=2)
connections <- list(
c(12,2,11,3), # for 1
c(1,12,13,4,3), # for 2
c(2,4,1,5), # for 3
c(2,3,5,6,13), # for 4
c(4,6,3,7), # for 5
c(7,8,13,4,5), # for 6
c(8,6,5,9), # for 7
c(9,10,13,6,7), # for 8
c(8,10,11,7), # for 9
c(11,12,13,8,9), # for 10
c(10,12,9,1), # for 11
c(11,1,2,13,10), # for 12
c(2,4,6,8,10,12) # for 13
# Matrix for full node count
fnodecount <- c(sides=2, matches=0)
# Looking at possible triangles given our new connections:
sides <- 3; ret.ls <- list()
for (i in 1:13) nodesearch(i, connections, sides, collapser)
(fnodecount <- rbind(fnodecount, c(sides, length(ret.ls))))
# We get 92 possible triangles
par(mfrow=c(10,10), mar=rep(0,4))
for (i in ret.ls) {
plot.node(seq(1,11,2), col=gray(1), border=gray(.7), ntext=F, clear=T)
plot.node(1:12, col=gray(1), border=gray(.7), ntext=F, clear=F)
plot.node(i, ntext=F, clear=F)
# I notice that plot row 2 col 1 and row 4 col 2 seems
# to be lines which means that there is some kind of error
# in the code somewhere.
# It is important to note that the more nodes/connections the
# more computationally complex the problem becomes. If there
# were no node elimination and each node had the same number of
# connections then the computations (c) would be approximately
# equal to:
# Computations = k*c^s
# Where s is the number of sides and k is the number of computations
# within each search algorithm.
# With the star since the average number of connections is
# 3.7 for a triangle an
# approximation of the number of computations is
# k*3.7^4=187k
# However, adding the six connections increases the average number
# of connections by 6/13.
# k*(3.7+6/13)^4=300k
# Thus even adding relatively few connections significantly increases
# the processing time.
# Going to a quadralateral
# k*3.7^5=693k
# With the new
# k*(3.7+6/13)^5=1248k
# This number is only an approximation because each iteration does
# decrease the number of possible connections. However, each iteration
# also results in a longer list of matches that need to be check through
# before adding to the list of solutions, k()'>0.
sides <- 4; ret.ls <- list()
for (i in 1:13) nodesearch(i, connections, sides, collapser)
(fnodecount <- rbind(fnodecount, c(sides, length(ret.ls))))
# Yeilds 351 possible matches | {"url":"https://www.r-bloggers.com/2014/03/the-star-puzzle/","timestamp":"2024-11-10T12:40:58Z","content_type":"text/html","content_length":"122289","record_id":"<urn:uuid:8d01e152-8ee7-4f0f-83ec-a73957c4c00c>","cc-path":"CC-MAIN-2024-46/segments/1730477028186.38/warc/CC-MAIN-20241110103354-20241110133354-00597.warc.gz"} |
Average Calculator - Symbolab
About Average Calculator
• An average calculator is a tool used to determine the average of a set of numbers. As a general concept in mathematics, an average refers to the central or typical value within a set of numbers.
The average is often considered a measure of central tendency, providing an overall summary of a broader dataset.
• Typically, the average calculator initiates a process of summing up all values within a dataset and then dividing it by the quantity of numbers in that set. For instance, the average of five,
ten, and fifteen will be the sum (5+10+15) divided by the number of items, which is three – resulting in an average of ten.
• A complex dataset may challenge individuals to calculate averages manually, which prompts them to resort to an average calculator. This tool provides quick, accurate results, hence saving time
and eliminating any risks of human error.
• Average calculators are handy, digital instruments which can range from being simple online tools to dedicated devices that perform calculations or functions beyond averaging numbers. They are
available in different formats, sizes, applications, and complexity.
• Some online calculators are designed to compute just the basic arithmetic mean or average, while others cater to various statistical averages such as weighted averages or moving averages. In
addition to calculating averages, some of these calculators also estimate other statistical measures such as variance, median, mode, and standard deviation, offering broader insights into data
• Users can benefit from an average calculator in many scenarios: students require it for statistical assignments, researchers use it to analyze experimental data, marketers deploy it to evaluate
sales and marketing trends, and economists utilize it for economic modeling.
• Typically, to use an average calculator, users input their set of numbers into the calculator, usually separated by commas or spaces. After that, they select the type of average they need to
compute (mean, median, mode, or range) and then command the calculator to execute the calculation. Within seconds, the calculator provides the average result.
• In many instances, these calculators would additionally display step-by-step instructions showing the complete computation process. This feature aids in understanding how the calculations work,
thereby assisting in learning and comprehension, especially for learners or novices.
• Notably, the fact an average calculator offers some advantages doesn’t mean it replaces analytical skills. Users still need to understand key statistical concepts and interpret the results
meaningfully. Without contextual knowledge and interpretation capabilities, raw averages generated from the calculator may not provide significant insights.
• In conclusion, whether being used to help with homework, making sense of survey results, or tracking financial market changes, an average calculator represents a useful tool for processing and
analyzing large datasets. By doing so, it contributes to streamlining decision-making processes, transforming large volumes of data into actionable insights. However, it is up to the users to
interpret and align these results to their objectives, questions, or hypotheses. After all, the calculator is only as intelligent as its user. | {"url":"https://ru.symbolab.com/calculator/statistics/average","timestamp":"2024-11-04T21:56:39Z","content_type":"text/html","content_length":"161441","record_id":"<urn:uuid:eec6cf5a-584c-48bb-ab5c-d118a096fe5a>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00270.warc.gz"} |
How to use GPU acceleration to solve linear equation Ax=b
In my experiment,I want to solve the equation Ax=b on GPU to accelerate the process.I have tried the iterative solver from Kryvol.jl and cuSOLVER,but the GPU is still slower than CPU.Is that normal?
using CUDA
using SparseArrays
using LinearAlgebra
using Krylov
using BenchmarkTools
using JLD2
# Function： Using BiCGStab on CPU to solve the sparse matrix linear equation
function solve_sparse_cpu(A, b)
x = zeros(Float32, size(b))
x=Krylov.bicgstab(A, b)
return x
# Function： Using BiCGStab on GPU to solve the sparse matrix linear equation
function solve_sparse_gpu(d_A, d_b)
d_x = CUDA.zeros(Float32, size(d_b))
d_x=Krylov.bicgstab(d_A, d_b, d_x)
return d_x
@load "C:/Users/13733/Desktop/DistributionPowerFlow/J.jld2" J
@load "C:/Users/13733/Desktop/DistributionPowerFlow/F.jld2" F
matrix_sizes = size(J,1)
for n in matrix_sizes
println("The scale of matrix A: $n x $n")
A = Float32.(sparse(J))
b = Float32.(-F)
d_A = CUDA.CUSPARSE.CuSparseMatrixCSC(A)
d_b = CuArray(b)
# CPU time consumption
cpu_time = @belapsed solve_sparse_cpu($A, $b)
println("CPU time consumption $cpu_time seconds")
# GPU time consumption
gpu_time = @belapsed solve_sparse_gpu($d_A, $d_b)
println("GPU time consumption: $gpu_time seconds")
println("speed up (CPU / GPU): $(cpu_time / gpu_time)")
The number of non-zero element in matrix J is 23958 and the size of J is 3498*3498.The output result is as follows.
The scale of matrix A: 3498 x 3498
CPU time consumption 0.1821423 seconds
GPU time consumption: 1.2135631 seconds
speed up (CPU / GPU): 0.15008885817309375
I found that in highly sparse matrices, solving on GPU is not as fast as on CPU
1 Like
How big is your system?
1 Like
A is a 3498*3498 sparse matrix
My GPU is NVIDIA GeForce RTX 4060 laptop,and my CPU is i9-12900HX
Hi @automaticvehiclerook!
For this kind of small linear systems, the GPU could be less efficient than CPU.
You could try a few things to improve performances:
• Use the CuSparseMatrixCSR format, it’s faster if you only need products with A and not its adjoint A'.
• Preallocate the workspace with solver = BicgstabSolver(d_A, d_b) and call the method in-place with bicgstab!(solver, d_A, d_b) to check if the majority of time spent is not related to
• Use the package KrylovPreconditioners.jl to create an operator optimized for multiple products A * v. You just need to pass op_A = KrylovOperator(d_A) to bicgstab or bicgstab!.
I also suggest to use CUDA.@profile to check what is the most expensive operation on GPU for your linear system.
Note that you can also have a faster code on CPU if you use multithreaded matrix-vector products and optimized BLAS.
Just adding using MKL, MKLSparse before calling a Krylov solver can lead to a nice speed-up.
The question is about GPU, but it’s better to have a good baseline on CPU for comparison.
2 Likes
I suggest to also try CUDSS.jl.
It’s an interface to sparse direct methods (LDL’, LL’ or LU) on NVIDIA GPUs.
It’s still a preview library of NVIDIA but it works well.
It should be more efficient than Krylov.jl for small problems.
4 Likes
Thank you for your reply！In fact, what I want to solve is a large-scale linear equation.I tested a 17036*17036 system and it took about 20 seconds on the CPU and 10 seconds on the GPU.I want to use
a preconditioner for Bicgstab but I don’t know how to do it on GPU.
I have to adjust the precision to Float64 to meet my engineering requirement
using CUDA
using SparseArrays
using LinearAlgebra
using Krylov
using BenchmarkTools
using JLD2
# Function： Using BiCGStab on CPU to solve the sparse matrix linear equation
function solve_sparse_cpu(A, b)
x = zeros(Float64, size(b))
x=Krylov.bicgstab(A, b)
return x
# Function： Using BiCGStab on GPU to solve the sparse matrix linear equation
function solve_sparse_gpu(d_A, d_b)
d_x = CUDA.zeros(Float64, size(d_b))
d_x=Krylov.bicgstab(d_A, d_b, d_x)
return d_x
@load "C:/Users/13733/Desktop/DistributionPowerFlow/J9241" J9241
@load "C:/Users/13733/Desktop/DistributionPowerFlow/F9241" F9241
matrix_sizes = size(J,1)
for n in matrix_sizes
println("The scale of matrix A: $n x $n")
A = Float64.(sparse(J))
b = Float64.(-F)
d_A = CUDA.CUSPARSE.CuSparseMatrixCSC(A)
d_b = CuArray(b)
# CPU time consumption
cpu_time = @belapsed solve_sparse_cpu($A, $b)
println("CPU time consumption $cpu_time seconds")
# GPU time consumption
gpu_time = @belapsed solve_sparse_gpu($d_A, $d_b)
println("GPU time consumption: $gpu_time seconds")
println("speed up (CPU / GPU): $(cpu_time / gpu_time)")
I give a few examples in the documentation of Krylov.jl:
On GPU, I suggest the following preconditioners:
• Jacobi / block-Jacobi
• Incomplete Cholesky with zero fill-in – IC(0)
• Incomplete LU with zero fill-in – ILU(0)
All of them are available in KrylovPreconditioners.jl.
You can find some examples here of how to use kp_ic0, kp_il0 and BlockJacobiPreconditioner on GPUs.
3 Likes
Thank you very much for your help. I will try it | {"url":"https://discourse.julialang.org/t/how-to-use-gpu-acceleration-to-solve-linear-equation-ax-b/120841","timestamp":"2024-11-14T08:50:12Z","content_type":"text/html","content_length":"39379","record_id":"<urn:uuid:1a6eb5cf-2e82-4195-b870-8cadafe90753>","cc-path":"CC-MAIN-2024-46/segments/1730477028545.2/warc/CC-MAIN-20241114062951-20241114092951-00487.warc.gz"} |
An Introduction to Python memorisation using functools
In Python you can trade time-complexity for an increase in the memory usage of a Python script. For example, the factorial function is defined recursively f(5) = 5! = 5 * 4 * 3 * 2 * 1, where 1 is
the base-case. But, what happens if we remember what 4! is equal to, by definition 5! is simply 5 * memo{4}
You could just use a Python dictionary like so:
memo = {0:1}
def factorial(n):
if n < 2:
return memo[0]
if n not in memo:
memo[n] = n * factorial(n-1)
return memo[n]
Python already has this built-in by importing functools from the standard library. Here’s the Fibonacci algorithm using LRU cache:
from functools import lru_cache
def fib(n):
if n < 2:
return n
return fib(n-1) + fib(n-2)
This just goes to show it is still possible to write performant functions that are easier to understand than there iterative equivalent. | {"url":"https://www.lewis8s.codes/python/algorithms/2019/12/29/python-memorisation-using-functools","timestamp":"2024-11-07T17:28:57Z","content_type":"text/html","content_length":"8474","record_id":"<urn:uuid:6cbdd3f1-6faf-47a3-8038-60e65d0ac344>","cc-path":"CC-MAIN-2024-46/segments/1730477028000.52/warc/CC-MAIN-20241107150153-20241107180153-00681.warc.gz"} |
Two features of the GINAR(1) process and their impact on the run length performance of geometric control charts
Morais, M. C.
Discrete-Valued Time Series, (2024), 179-190
The geometric first-order integer-valued autoregressive process (GINAR(1)) can be particularly useful to model relevant discrete-valued time series, namely in statistical process control. We resort
to stochastic ordering to prove that the GINAR(1) process is a discrete-time Markov chain governed by a totally positive of order 2 (TP2) transition matrix. Stochastic ordering is also used to
compare transition matrices referring to pairs of GINAR(1) processes with different values of the marginal mean. We assess and illustrate the implications of these two stochastic ordering results,
namely on the properties of the run length of geometric charts for monitoring GINAR(1) counts. | {"url":"https://cemat.tecnico.ulisboa.pt/document.php?project_id=5&member_id=90&doc_id=3641","timestamp":"2024-11-05T10:22:09Z","content_type":"text/html","content_length":"8778","record_id":"<urn:uuid:2ae3d7ba-569d-43ec-bef2-fdc6204399fb>","cc-path":"CC-MAIN-2024-46/segments/1730477027878.78/warc/CC-MAIN-20241105083140-20241105113140-00711.warc.gz"} |
Learn Google Sheet Formulas The Hard Way
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Learn Google Sheet Formulas The Hard Way
About this Tutorial
One weird way to learn more about what Google Sheets can do. https://bettersheets.co/formulas
Video Transcript
0:00 <affirmative>. Hey, so I understand that Google Sheets is pretty darn hard. It's starting from a blank slate every single time.
0:09 It's literally white and black and it's just staring at you in the face with nothing. No, no, nothing, no sense of soul behind those eyes.
0:19 Well, here's something that I wanna share with you that I don't think many people will do. So that's why I called this Learn Google Formulas.
0:30 The hard way, at least Google Sheets formulas. We should probably fix that. Google Sheets formulas. All right, this is why I called it Google Sheets formulas.
0:39 The hard way is that I don't think many people will do it. I don't think you'll necessarily do this, or you might start and you'll stop.
0:47 But I'm gonna give you, at the end of this video a trick or a hack a tool. Let's say that I've created a better sheets to help you, so don't worry.
0:58 What I'm about to show you is so stupidly simple as well, that you might not do it and you might not think you'll get enough out of it.
1:06 But I can tell you unequivocally, this is very much how I learned a lot of Google Sheet formulas is you can go into any sheet.
1:15 We just started a new sheet sheet.new, and I type in any cell. Are you an A one or are you a B two?
1:23 I'm a B two myself. Starter here. Just type the equal sign. And once you type the equal sign and then you do any letter, you can do A, you can do B, you can do C, D, E, F, all the way up to z
<laugh>, I think Y as well, and X.
1:40 Yeah, I think there's one for every, there's a formula for every letter, at least one or two. It's a Q.
1:45 Yeah, they're definitely q u. There's definitely for every 20, for all 26 letters, there is a formula. Now, why is this important?
1:54 Why would you start at a or why would you start it Maybe Doo Z down to a check out the list, see what they are, and click on them.
2:02 And if you click on them and you click on details, you can read a little short description about it. You can see what you need to, what syntax it is right here.
2:13 What is the kind of variables it does? What is it? Return? You can also click here to learn more. For some reason, Michael Cogen gives me an error every time I do this.
2:24 It gives you general usage. It's some examples, and then asks you if this was helpful or not. Now, here's the thing.
2:31 Google has been very good by providing this information. We can open a new window. Actually Google Docs editor help, but it's very difficult to really understand how this is used, like the z
2:49 And we can go through each and every one. I can, I have to reload this sheet every time. Go back to B two equals A.
2:57 And there's something like abs. Let's, well, let's let it load up. It takes a moment here. Abs there absolute number of absolute value of a number.
3:08 Now why would you wanna like, look through all these? Well, there's like two ways that we learn at least two binary ways.
3:17 I, and I know there's ways to learn by listening. There's ways to, by to learn, by seeing you experiential. There's diff that's one type of way types, buckets of learning.
3:28 But I think there's fundamentally two ways to learn thi something, especially Google Sheets and Google Sheets formulas and getting better.
3:35 There's two ways. One top down where you have some ultimate goal and you just need to get through it, and you need to know, how do I do the thing that I'm doing?
3:44 And then the other way is bottom up understanding what is capable, what Google Sheets is capable of. Now, it is a much slower, it is a much less certain way of learning, right?
4:00 If you just go for the thing you're trying to do and know, okay, probably 99% of the time, if I'm trying to do something, someone else in the last 15 years has tried to do this in Google Sheets.
4:12 So I'll just search for that. I'll Google for that. You can also learn, and, and that's great. You'll learn that way.
4:19 And, and, and that's many <laugh> of the ways to learn in better sheets. A lot of them, my videos are just a tutorial showing you how to do something, and then you end up learning some cool
little formulas on the way.
4:29 But here, I'm trying to tell you, there is this crazy other way, which is you can read every one of these formulas.
4:37 Do you know how many there are? There's 502. I know that because here comes the trick, here comes the tool.
4:45 If you truly go to a sheet and start doing equals great <laugh>, but also you might be a great candidate for joining Better Sheets.
4:55 And if you're already a, a member, member of Better Sheets, then you'll know that if you go to Better sheets.co, let's just go to Better sheets.co.
5:07 Up at the top is a link for formulas. It is open and available to everyone, member or not member. So if you're not a member watching this, go ahead to Better sheets.co/formulas.
5:18 And here are 500, actually 501. Google Sheets formula is everything from ABS to Z test and everything in between. We have, if let's find, if here I have it a couple times.
5:33 There's a few ifs. There's, if, here's the biggest point, I don't know where to move my face. This is the biggest point though.
5:42 It has very similar information as Google, but it also has links to any and all better sheet videos that feature that formula.
5:52 So here we have 20 Better Sheet formulas better Sheets tutorials That include the formula of if, if is literally one of my favorite and most popular formulas.
6:06 I love it. It it's very useful in many, many use cases. You can also see these formulas if you go over to the tutorials tab.
6:13 But I wanted to show you this formulas up here tutorials, and you can filter all the tutorials by whatever f formula you're looking for.
6:22 But this way you can go and discover 501 formulas over at Better sheets.co/formulas. So there'll be a link in the description here for you, or just go better sheets.co/formulas or just go to
better sheets.co and click on formulas.
6:37 And here's every single formula. It's really fun. I'm gonna be adding more information as we go Right now. We have a link to Google Docs.
6:44 We have links to any and all sheet videos that Better Sheets features. Some are free, some are paid. You'll, if you are not a member, you'll see a lock on the videos if you are a member.
6:59 And you can track which videos you're watching. So if you have already watched the video, it'll have a check mark.
7:04 Mark. If you haven't watched the video, it, it'll have a number. Then you can go to the video, watch it.
7:09 Oops. So you can go to a video and you can see, watch the video, and then you can click Mark is completed.
7:14 So, you know, you watch that video. And so you can watch, re-watch video in different lenses in different context if you're looking for more information about that formula.
7:23 Again, all of the, all the videos that are listed here are not gonna feature, like, they're not going to teach you how to use the formula.
7:31 They're just gonna show you how to use the formula in that video. The form, the vi <laugh>, sorry, the formula will appear at some point in that video.
7:42 So, just a caveat, I hope you enjoy this, and I do think that there is a great opportunity to learn a lot about Google Sheet formulas by just going through A, B, C, D.
7:55 But I try to make this easier for you and learn a lot of different ones, and you'll be able to learn the most common ones that we use in videos.
8:02 Sometimes there won't be any videos in better sheets for some of these, like the, the very weird and esoteric ones that nobody uses or very few people use.
8:11 But we'll be adding more and more videos, go on better sheets all the time. So keep become a member, stay notified of videos coming out.
8:20 And I hope you can learn Google Sheets an easy way or by working on the projects you wanna work on.
8:26 Or go ahead, learn Google Sheet formulas the hard way by just going through and reading all of them. There's some pretty interesting ones that I was like, whoa, that exists.
8:36 You can do that. You can do that in Google Sheets. So enjoy. Bye. | {"url":"https://bettersheets.co/tutorials/learn-google-sheet-formulas-the-hard-way","timestamp":"2024-11-05T07:23:13Z","content_type":"text/html","content_length":"36542","record_id":"<urn:uuid:45a0fe13-8701-4749-af06-24917bc5a316>","cc-path":"CC-MAIN-2024-46/segments/1730477027871.46/warc/CC-MAIN-20241105052136-20241105082136-00099.warc.gz"} |
Dynamic analysis of reciprocating compressor system with translational clearance and time-varying load
Dynamic behavior of reciprocating compressor system, with translational clearance between the crosshead and guide under time-varying cylinder load, is investigated. In order to analyze the dynamic
response of the system with translational clearance, a novel nonlinear dynamic model is established based on the Lagrangian approach. The numerical solution of the dynamic equation is calculated by
the Runge-Kutta method. The results show that the translational clearance has a great effect on the reciprocating compressor, and the more the translational clearance, the great the influence.
Moreover, the phase space of the crosshead reveals that the reciprocating compressor system with translational clearance has chaotic characteristics.
1. Introduction
Reciprocating compressors are one of the most popular machines used in petroleum and chemical production processes, such as gas compression, petroleum transportation and natural gas transportation
[1, 2]. In practice, with reciprocating compressor working on a period of time, as the result of manufacture tolerance and wear, some translational clearances in its joints commonly exist, and they
are inevitable. In the case of oversized joint clearances, contact forces generate impulsive effect, and this situation causes increased vibration and noise, and reduce system reliability, stability,
life and precision. So, clearances play a significant role in the prediction of kinematic and dynamic behavior of reciprocating compressor [3].
In the last decade, the fault diagnosis researches of reciprocating compressor are focused on vibration signal extraction. On the valve failure, Wang et al. [4] presented an experimental study of the
fault diagnosis of reciprocating compressor valves with acoustic emission technology and simulated valve motion. In their study, the results indicate that an earlier occurrence of the suction process
can diagnose suction valve leakage and that an earlier occurrence of the discharge process can be used for detecting discharge valve leakage. In addition, some meaningful research results have been
made, such as the characteristics extraction of the piston rod [5], the characteristic extraction of the impact signal at the reverse angle [6], and so on. However, few scholars have studied the
failure mechanism of reciprocating compressor with translational and revolute clearances fault. Zhao et al. [7] performed a parameter optimization approach for planar joint clearance model and its
application for dynamics simulation of reciprocating compressor. The dynamics response experimental test verified the effectiveness of this application. Jiang et al. [8] focused on the study of the
dynamic response and diagnosis method on wear fault of small-end bush of a connecting rod based on the dynamic simulation and vibration signal analysis.
Because the single cylinder reciprocating compressor is a crank slider mechanism shown in Fig. 1, the dynamics analysis of the crank slider mechanism with translational clearance can be used for
reference. Flores et al. [9, 10] developed a methodology for a dynamic modeling and analysis of rigid multibody systems with translational clearance joints based on the non-smooth dynamics approach.
Zhuang and Wang [11] carried out a modeling and simulation method for the rigid multibody system with frictional translational joints between a slider and guide.
In this work, based on the previous research results, we carry out the dynamic analysis of the reciprocating compressor with translational clearance fault under time-varying cylinder load. The paper
is organized as follows: The dynamic model of single cylinder reciprocating compressor with translational clearance fault is established in Section 2. The influence of translational clearance size is
discussed in Section 3. Furthermore, the conclusions of this paper are given in Section 4.
Fig. 1Schematic diagrams of single cylinder reciprocating compressor
2. Dynamic model of single cylinder reciprocating compressor with translational clearance
2.1. Model of time-varying cylinder load
The crankshaft of single cylinder reciprocating compressor rotates a circle, driving the connecting rod, the crosshead and the piston to reciprocate. The cylinder can realize the four processes of
expansion, suction, compression and exhaust. In a circle, As the crankshaft turns clockwise, the crosshead and piston move from right to left, working volume that is located at the right side of the
piston gradually increases, cavity gas gradually expands, and cylinder pressure gradually decreases, which is expansion process. When the cylinder internal pressure decreases to slightly less than
the external pressure of the cavity, the inlet valve is opened, until the crosshead and piston move to the far left, which is suction process. In the suction process, cylinder pressure is almost
constant. The suction process is accomplished at the far left. The compression and exhaust processes are mostly the opposite of expansion and suction. It is obvious that the cylinder pressure can be
seen as a time-varying load. Since the change of cylinder pressure is periodic, it is more appropriate to use the crankshaft rotation angle as the variable for the cylinder pressure expression.
Pressure expression can be expressed as follow:
where $P$ is the cylinder pressure, ${P}_{s}$ denotes cylinder pressure coefficient, $\mu$ is given by:
$\mu =\left\{\begin{array}{l}\mathrm{s}\mathrm{i}\mathrm{n}\left(\frac{\pi }{2}-{\theta }_{1}\right),\mathrm{}\mathrm{}\mathrm{}0°+2n\pi \le {\theta }_{1}\le \frac{7\pi }{16}+2n\pi \left(n\mathrm{}\
text{is integer}\right),\mathrm{}\mathrm{}\text{Expansion}\text{,}\\ \mathrm{s}\mathrm{i}\mathrm{n}\frac{\pi }{16},\mathrm{}\mathrm{}\mathrm{}\frac{7\pi }{16}+2n\pi \le {\theta }_{1}\le \pi +2n\pi \
left(n\text{is integer}\right),\mathrm{}\mathrm{}\text{Suction}\text{,}\\ -\mathrm{s}\mathrm{i}\mathrm{n}\left({\theta }_{1}+\frac{\pi }{16}\right),\mathrm{}\mathrm{}\mathrm{}\pi +2n\pi \le {\theta }
_{1}\le \frac{23\pi }{16}+2n\pi \left(n\mathrm{}\text{is integer}\right),\mathrm{}\mathrm{}\text{Compression}\text{,}\\ 1,\mathrm{}\mathrm{}\mathrm{}\frac{23\pi }{16}+2n\pi \le {\theta }_{1}\le 2\pi
+2n\pi \left(n\mathrm{}\text{is integer}\right),\mathrm{}\mathrm{}\text{Exhaust}.\end{array}\right\\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}$
where ${\theta }_{1}$ denotes the crankshaft rotation angle in a clockwise direction, $n$ is the number of cycles of crankshaft rotation.
2.2. Contact model with translational clearance
The translational joint is obtained through an infinitely enlarged translational joint radius. Therefore, the contact model with translational clearance can refer to the revolute clearance. When the
translational clearance shown in Fig. 2 is too large, impact will take place between the crosshead and guide when the following condition is met at about the relative penetration depth of $\delta$.
The relative penetration depth of $\delta$ can be defined by:
$\delta =\left|{y}_{3}\right|-{r}_{c},$
in which $\delta$ denotes the relative penetration depth, ${y}_{3}$ is the centroid coordinates of the crosshead in the y direction, ${r}_{c}$ represents translational clearance size.
Assuming that the rotation of the crosshead can be ignored during the collision, namely, the crosshead is only perpendicular to the slide when the collision appears. According to Lankarani-Nikravesh
contact force model and Ambrósio friction model, the contact force ${Q}_{c}$ can be written as:
${Q}_{c}=K{\delta }^{m}\left(1+{{c}_{f}}^{2}{{c}_{d}}^{2}{\right)}^{\frac{1}{2}}\left[1+\frac{3\left(1-{{c}_{r}}^{2}\right)}{4}\frac{\stackrel{˙}{\delta }}{{\stackrel{˙}{\delta }}^{\left(-\right)}}\
right],\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\psi =\alpha +\phi ,$
where ${c}_{f}$ is the dynamic friction coefficient; ${v}_{t}$ is the relative tangential velocity along the direction of the guide; ${c}_{d}$ is a dynamic correction coefficient; $m$ is the
nonlinear power exponent, which is generally set to 1.5 for metallic surfaces; ${c}_{r}$ denotes the restitution coefficient; $\stackrel{˙}{\delta }$ is the relative penetration velocity; ${\stackrel
{˙}{\delta }}^{\left(-\right)}$ is the initial impact velocity of the impact point, which should be updated for each impact process; $K$ is the stiffness coefficient; $\psi$ represent direction of
the force ${Q}_{c}$; $\alpha$ and $\phi$ are computed by the following equation:
$\phi =\mathrm{a}\mathrm{r}\mathrm{t}\mathrm{a}\mathrm{n}\frac{{F}_{t}}{{F}_{n}},$$\alpha =\left\{\begin{array}{l}\frac{\pi }{2},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm
{}\mathrm{}\text{Impact with the upper surface of the slide}\text{,}\\ -\frac{\pi }{2},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{I}\text{mpact with the lower surface of the slide}\text{,}\
end{array}\right\\mathrm{}\mathrm{}\mathrm{}\mathrm{}\text{or}\mathrm{}\mathrm{}\alpha =\frac{\pi }{2}\cdot \mathrm{s}\mathrm{i}\mathrm{g}\mathrm{n}\left({y}_{3}\right).$
Fig. 2Schematic of single cylinder reciprocating compressor
2.3. Model of dynamics
Distinctly, there are two degrees of freedom for reciprocating compressor systems with translational clearance. So, two generalized coordinates can be represented by ${\theta }_{1}$ and ${\theta }_
{2}$. As can be seen from Fig. 2, the velocity of crankshaft, connecting rod and crosshead can be obtained as follow, respectively:
${\stackrel{˙}{x}}_{1}=-\frac{1}{2}{l}_{1}{\stackrel{˙}{\theta }}_{1}\mathrm{s}\mathrm{i}\mathrm{n}{\theta }_{1},\mathrm{}\mathrm{}\mathrm{}\mathrm{}{\stackrel{˙}{y}}_{1}=\frac{1}{2}{l}_{1}{\stackrel
{˙}{\theta }}_{1}\mathrm{c}\mathrm{o}\mathrm{s}{\theta }_{1},$
${\stackrel{˙}{x}}_{2}=-{l}_{1}{\stackrel{˙}{\theta }}_{1}\mathrm{s}\mathrm{i}\mathrm{n}{\theta }_{1}-\frac{1}{2}{l}_{2}{\stackrel{˙}{\theta }}_{2}\mathrm{s}\mathrm{i}\mathrm{n}{\theta }_{2},\mathrm
{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}{\stackrel{˙}{y}}_{2}={l}_{1}{\stackrel{˙}{\theta }}_{1}\mathrm{c}\mathrm{o}\mathrm{s}{\theta }_{1}+\frac{1}{2}{l}_{2}{\stackrel{˙}{\theta }}_{2}\mathrm{c}\
mathrm{o}\mathrm{s}{\theta }_{2},$
${\stackrel{˙}{x}}_{3}=-{l}_{1}{\stackrel{˙}{\theta }}_{1}\mathrm{s}\mathrm{i}\mathrm{n}{\theta }_{1}-{l}_{2}{\stackrel{˙}{\theta }}_{2}\mathrm{s}\mathrm{i}\mathrm{n}{\theta }_{2},\mathrm{}\mathrm{}\
mathrm{}\mathrm{}{\stackrel{˙}{y}}_{3}={l}_{1}{\stackrel{˙}{\theta }}_{1}\mathrm{c}\mathrm{o}\mathrm{s}{\theta }_{1}+{l}_{2}{\stackrel{˙}{\theta }}_{2}\mathrm{c}\mathrm{o}\mathrm{s}{\theta }_{2},$
where ${\stackrel{˙}{x}}_{i}$ and ${\stackrel{˙}{y}}_{i}$ ($i=$1, 2, 3) are the velocity of crankshaft, connecting rod and crosshead, respectively; ${\theta }_{1}$ and ${\theta }_{2}$ are the angle
of the crankshaft and connecting rod with the $x$-axis, respectively. ${l}_{1}$ and ${l}_{2}$ represent the length of crankshaft and connecting rod, respectively.
According to Eqs. (6)-(8), the kinetic energy and potential energy of crankshaft, connecting rod and crosshead can be calculated as follows:
${E}_{1}=\frac{1}{6}{m}_{1}\left({l}_{1}{\stackrel{˙}{\theta }}_{1}{\right)}^{2},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}{E}_{2}==\frac{1}{2}{m}_{2}\left({l}_{1}{\stackrel{˙}{\theta }}_{1}{\
right)}^{2}+2{m}_{2}{J}_{2}{\left({l}_{2}{\stackrel{˙}{\theta }}_{2}\right)}^{2}+\frac{1}{2}{m}_{2}{l}_{1}{l}_{2}{\stackrel{˙}{\theta }}_{1}{\stackrel{˙}{\theta }}_{2}\mathrm{c}\mathrm{o}\mathrm{s}\
left({\theta }_{2}-{\theta }_{1}\right),$${E}_{3}=\frac{1}{2}{m}_{3}\left({l}_{1}{\stackrel{˙}{\theta }}_{1}{\right)}^{2}+\frac{1}{2}{m}_{3}\left({l}_{2}{\stackrel{˙}{\theta }}_{2}{\right)}^{2}+{m}_
{3}{l}_{1}{l}_{2}{\stackrel{˙}{\theta }}_{1}{\stackrel{˙}{\theta }}_{2}\mathrm{c}\mathrm{o}\mathrm{s}\left({\theta }_{2}-{\theta }_{1}\right),$
$E={E}_{1}+{E}_{2}+{E}_{3},\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}\mathrm{}V=\left(\frac{1}{2}{m}_{1}+{m}_{2}+{m}_{3}\right){l}_{1}g\mathrm{s}\mathrm{i}\mathrm{n}{\theta }_{1}+\left(\frac{1}{2}
{m}_{2}+{m}_{3}\right){l}_{2}g\mathrm{s}\mathrm{i}\mathrm{n}{\theta }_{2},$
where, ${E}_{1}$, ${E}_{2}$ and ${E}_{3}$ are the kinetic energy of crankshaft, connecting rod and crosshead, respectively. $V$ is the sum of potential energy for the reciprocating compressor system.
${m}_{1}$, ${m}_{2}$ and ${m}_{3}$ are the mass of crankshaft, connecting rod and crosshead, respectively.
Substitute Eqs. (1), (4) and (10) into following Lagrange motion equation Eq. (11), the dynamic equation can be obtained:
$\frac{d}{dt}\left(\frac{\partial E}{\partial {\stackrel{˙}{q}}_{j}}\right)-\frac{\partial E}{\partial {q}_{j}}+\frac{\partial U}{\partial {q}_{j}}={Q}_{j},\left(j=1,\mathrm{}2\right),$
where $E$ and $U$ are the kinetic and potential energies of the reciprocating compressor system, respectively. ${Q}_{j}$ is the nonconservative generalized force corresponding to the generalized
coordinate. The expression of ${Q}_{j}$ and $L$ are given as:
${Q}_{j}={\sum }_{i=1}^{3}{{\stackrel{\to }{F}}_{i}}^{\mathrm{*}}\frac{\partial {\stackrel{\to }{V}}_{i}}{\partial {\stackrel{˙}{q}}_{j}}+{{\stackrel{\to }{M}}^{\mathrm{*}}}_{i}\frac{\partial {\
stackrel{\to }{\omega }}_{i}}{\partial {\stackrel{˙}{q}}_{j}},\mathrm{}\mathrm{}\mathrm{}\mathrm{}L=E-U,$
where ${{\stackrel{\to }{F}}_{i}}^{\mathrm{*}}$denotes the resultant of external force acting at the center of mass. ${{\stackrel{\to }{M}}^{\mathrm{*}}}_{i}$ represents the external torque acting on
body $i$. ${\stackrel{\to }{V}}_{i}$ and ${\stackrel{\to }{\omega }}_{i}$ denote the translational and rotational velocity for the mass center of body $i$, respectively.
3. Results and discussion
In this section, we discuss the dynamic behavior of reciprocating compressor with translational clearance between the crosshead and guide. Solving Eq. (11) by Runge-Kutta Method, the numerical
solution of ${\theta }_{1}$ and ${\theta }_{2}$ can be obtained. Subsequently, the corresponding numerical solution of displacement, velocity and acceleration of the crosshead can be calculated. In
the numerical solution, 2D12 reciprocating compressor is used for the research object, and its structural parameters are as follows: ${l}_{1}=$0.12 m, ${l}_{2}=$0.6 m, ${m}_{1}=$1 kg, ${m}_{2}=$5 kg,
${m}_{3}=$1 kg, ${c}_{f}=$0.5, ${c}_{r}=$0.9, ${P}_{s}=$2×10^5, $K=$2.399×10^10. Figs. 3 to 5 display the dynamic response results of reciprocating compress with different translational clearance in
y direction under time-varying cylinder load.
As can be seen from Figs. 3 to 5, with the increase of translational clearance, the influences of the crosshead increase in $y$ direction. It is noteworthy that the variational clearance does not
influence the crosshead displacement in a conspicuous way. When the clearance sizes increase from 0.1 mm, 0.2 mm to 0.3 mm, the corresponding maximal deviation value of the displacement increases
from the 0.1519 mm, 0.1522 mm to 0.1523 mm at the low dead point, and the maximal deviation value of velocity increases from 0.0056 m/s, 0.1493 m/s to 0.2297 m/s. In sharp contrast, the crosshead
acceleration is distinctly influenced and the maximal peak value of crosshead acceleration increases from 1455 m/s^2, 2351 m/s^2 to 3190 m/s^2. Obviously, the more the translational clearance size,
the great the influence of displacement, velocity and acceleration of the crosshead. The influence of acceleration is larger than the displacement and velocity.
Fig. 3Dynamic response of reciprocating compress with translational clearance 0.1 mm in y direction: a) displacement response; b) velocity response; c) acceleration response; d) phase space
Fig. 4Dynamic response of reciprocating compress with translational clearance 0.2 mm in y direction: a) displacement response; b) velocity response; c) acceleration response; d) phase space
Fig. 5Dynamic response of reciprocating compress with translational clearance 0.3 mm in y direction: a) displacement response; b) velocity response; c) acceleration response; d) phase space
In addition, Fig. 3(d), Fig. 4(d) and Fig. 5(d) depict the phase trajectories of displacement and velocity. one can observe that the strange attractors are shown in Fig. 3(d), Fig. 4(d) and Fig. 5
(d). That is to say, the reciprocating compressor system indicates chaotic behavior.
4. Conclusions
In this interesting work, the dynamic behavior of reciprocating compressor for single cylinders is studied with translational clearance. The nonlinear dynamical equation is established under the
time-varying cylinder load, and the numerical solution of the equation is obtained by MATLAB software. By analyzing the dynamic response of the reciprocating compressor system with translational
clearance, some dynamic behaviors are obtained as follows:
1) With the increase of translational clearance, the influence of displacement, velocity and acceleration of the crosshead increase in $y$ direction, where the influence of displacement, velocity and
acceleration gradually increases.
2) The reciprocating compressor system with translational clearance can be observed strange attractors. the result reveals that this system is characterized by chaotic behavior with translational
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• Ma J., Jiang Z. N., Gao J. J. A reciprocating compressor fault diagnosis method based on piston rod axis orbit. Journal of Vibration Engineering, Vol. 25, Issue 4, 2012, p. 453-459, (in Chinese).
• Du X. Y., Jiang Z. N. Fault diagnosis of the small end bushing of connecting rod of reciprocating compressor based on angular domain. Compressor Technology, Vol. 5, 2012, p. 62-64, (in Chinese).
• Zhao H. Y., Xu M. Q., Wang J. D. A parameters optimization method for planar joint clearance model and its application for dynamic s simulation of reciprocating compressor. Journal of Sound and
Vibration, Vol. 344, 2015, p. 416-433.
• Jiang Z. N., Mao Z. W., Zhang Y. D. A study on dynamic response and diagnosis method of the wear on connecting rod bush. Journal of Failure Analysis and Prevention, Vol. 17, Issue 4, 2017, p.
• Flores P., Ambrósio J., Claro J.C., Lankarani H. M. Translational joints with clearance in rigid multibody systems. Journal of Computational and Nonlinear Dynamics of ASME, Vol. 3, Issue 1, 2008,
p. 110071.
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About this article
Mechanical vibrations and applications
reciprocating compressor
translational clearance
time-varying cylinder load
This paper was supported by the following research projects: “Fujian Natural Science Foundation” (Grant #2015J01643), “Ningde City Science and Technology Project” (Grant #20150034), “Education
Science Project of Young and Middle-aged Teachers of Universities in Fujian Province” (Grant #JZ160396 and Grant #JAT160527).
Copyright © 2017 JVE International Ltd.
This is an open access article distributed under the
Creative Commons Attribution License
, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. | {"url":"https://www.extrica.com/article/19331","timestamp":"2024-11-08T20:49:53Z","content_type":"text/html","content_length":"142348","record_id":"<urn:uuid:c194829b-d4be-4186-a5d2-f3bb1feaa7e5>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00207.warc.gz"} |
Problem Solving Reflexions Of A Mathematical Learner Essay
Unit Title : Problem Solving Reflexions Of A Mathematical Learner
Assignment Type : Essay
Word Count : 1600 words
1.Critically reflect on your own memories of being a mathematical learner. What beliefs, attitudes and emotions do you have about mathematics. What experiences have you had? How do you remember
learning? Connect this discussion back to theories about mathematical acquisition and learning.
Please refer chapter 3 matematic book attached AND use my experience.(ABOVE) you are more than welcome to change the wording but not the idea.
Problem Solving Reflexions Of A Mathematical Learner Essay
A lot of the time when we hear the term “problem-solving”, our first though is school mathematical exercises, and if we apply that concept to our lives, we associate “problem-solving” with a serie of
complicated events in our lives that we were able to resolve most of the times.
Problem-solving is an essential skill for everyone in the everyday life, and is not something that is adquired in the adulthood , since children we face different situations that required our logical
thinking, and this skill is build since our childhood.
Problem Solving Reflexions Of A Mathematical Learner Essay
From my earliest memories, mathematics was always a strong point. My father cultivated mathematics as something so important and funny that my sister and I loved doing maths in our early years. Easy
exercises like doing groceries were the most exciting plan. If we had the correct shopping price, we could pick up whatever we wanted from the shop, and of course, my sister and I were delighted with
this treat. Over the years, he continued teaching us the importance of mathematics and how can we use numbers for everything in our lives, from knowing the buses timetable and how much time do we
have to wait, even sharing was mathematically introduced to us, I will never forget how much my sister take advantage of my poor knowledge about fractions, she always got the most significant pieces
of cake, which encourage me to learn faster. By the age of 5, I already knew how to add, subtract, multiplicate and I was learning to divide. Connect with brofenbrenner
Problem Solving Reflexions Of A Mathematical Learner Essay
Unfortunately maths in year one (primary school) felt horrible. Everything needed to be learned by memory. So if I answer using a different process it does not matter if my answer was correct; my
teacher used to fail my home work. Situation that frustrated me profoundly, produced me stress and unnecessary anxiety. Connect with Piaget
Problem Solving Reflexions Of A Mathematical Learner Essay
On my second year my parents aware and concern for that situation solicitate to the principal to speak with my teacher and from there everything went way better.
PART B:
1. Identify and discuss how these beliefs, attitudes, and emotions impact in your teaching of mathematics in early childhood contexts. Why is it important to reflect on these beliefs? Relate this
with the learning out comes
PART C:
Identify and discuss how you going to build your content knowledge about mathematics. Name some activities and give some examples Identify professional organisations and/or professional development
opportunities that would be beneficial.
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July 2017
A recent viewer of my new Pluralsight course had a question about data gateways and Power BI Premium. Specifically, do you need a pro license to install and administer data gateways? The short answer
is probably not!
Installing data gateways
So when you install a data gateway, you need to log in as a user to register it with your tenant. Well it turns out that whoever is used there is set as the default admin for that gateway. I created
a user with just a power BI free license, and I was able to install and administer that gateway just fine. I was also able to assign it to other gateways that already existed.
So, for normal usage you don’t have to be licensed with pro to setup and configure data gateways. I was honestly a bit surprised by this, but in retrospect is makes sense. Pro licensing is all about
consuming reports.
What about Premium?
So, the original question was about Power BI Premium. Unfortunately, there’s no developer tier for me to test on, but I have a few guesses.
First, I reviewed the white paper and the distinction it makes between pro users and infrequent users is about producing versus consuming reports. It doesn’t really talk much about administration
from what I could tell. Same thing for the faq:
Do I need Power BI Pro to use Power BI Premium?
Yes. Power BI Pro is required to publish reports, share dashboards, collaborate with colleagues in workspaces and engage in other related activities.
Next, I did some searching, and found a page about capacity admins, but that doesn’t relate to data gateways specifically.
So based on what I found, I would assume that you don’t need a pro license to manage data gateways for premium. I would assume it would be a similar experience to normal Power BI.
New presentation: just enough database theory for Power BI
I just gave a presentation for the Excel BI virtual group on database theory, and I’m really happy with how it went. I think it’s an undeserved topic quite honestly. So many people in the excel
world learn everything ad-hoc and never have a chance to learn some of the fundamentals.
A number of questions came up relating to the engine and how the performance works. If you are interested in more detail on that, I suggest checking out my talk on DAX.
Here are the slides for my talk:
Just Enough Database Theory for PowerPivot 2017-20-2017
Video is coming soon as well.
Why DAX is a PITA: part 1
• So, I think that DAX is a pain in the butt to use and to learn. I talk about that in my intro to DAX presentation, but I think it boils down to the fact that you need a bunch of mental concepts
to have a proper mental model, to simulate what DAX will do. This is very deceptive, because it looks like Excel formulas on steroids, but conceptually it’s very different.
Here is the problem with DAX, in a nutshell:
This example below is a perfect example of that sharp rise in learning curve, and dealing with foreign concepts like calculated columns, measures, applied filters, and evaluation contexts.
So, one of the things I’m hoping to catalog are example where DAX is a giant pain if you don’t know what you are doing. People make it look really simple and smooth, and that can be frustrating
sometimes. Let’s see more failures!
How do I GROUPBY in DAX?
John Hohengarten asked me a question recently on the SQL Community Slack. He said:
I need to sum an amount column, grouped by a column
Measure 1 :=
GROUPBY (
“Total AR Amt Paid calc”, SUM ( det[amt] )
I’m getting a syntax error
So automatically, something seemed off to me. Measures are designed to return a single value, given the filter context that’s applied to them. That means you almost always need some aggregate
function at an outer level. But based on the name, you wouldn’t necessarily expect GROUPBY to return a single value. It would return values for each grouping instance.
If we take a look at the definition for GROUPBY(), we see it returns a table, which makes sense. But if you are new to DAX, this is really unintuitive because DAX works primarily in columns and
tables. This is a really hard mental shift, coming from SQL or Excel.
What do you really want?
None of this made any sense to me. Why would you try to put a GROUPBY in a measure? That’s like trying to return an entire table for a KPI on a dashboard. It just doesn’t make sense. So I asked John
what he was trying to do.
He sent me an image of some data he was working with. On the far left is the document id and on the far right is the transaction amount.
He wanted to add another column on the right, that summed up all of the amounts for transactions with the same document. In SQL, you’d probably do this using a Window function with a SUM aggregate,
like here.
Calculated columns versus measures
This highlights another piece of DAX that is unintuitive. You have two ways of adding business logic: calculated columns and measures. The both use DAX, both look similar and are added in slightly
different spots.
But semantically and technically, they are very different beasts. Calculated columns are ways of extending the table with new columns. They are very similar to persisted, computed columns in SQL. And
they don’t care at all about your filters or front-end, because the data is defined at time of creation or time of refresh. Everything in a calculated column is determined long before you are
interacting with them.
Measures on the other hand, are very different. They are kind of like custom aggregate functions, like if you could define your own version of SUM. But to carry the analogy, it would be like if you
had a version of SUM that could manipulate the filters you applied in your WHERE clause. It gets weird.
My point is, if you don’t grok the difference between calculated columns and measures, you will never be able to work your way around the problem. You will be forced to grope and stumble, like
someone crawling in the dark.
Filter context versus row context
So in this case we’ve determined we actually want to extend the table with a column, not create a free-floating measure. Now we run headlong into our next conceptual problem: evaluation contexts.
In DAX there are two types of evaluation contexts: row contexts and filter contexts. I won’t go too deep here, but they define what a formulas can “see” at any given time, and in DAX there are many
ways to manipulate these contexts. This is how a lot of the time intelligence stuff works in DAX.
In this case, because we are dealing with a calculated column, we have only a row context, not filter context. Essentially, the formula can only see stuff in the same row. Additionally, if we use an
aggregate like SUM, it only cares about the filter context. But the filter context comes from user interaction. Because this data is defined way before that, there is no filter context.
This is another area, where if you don’t understand these concepts you are SOL. Again, for the newbie, DAX is a pain.
What’s the solution?
So what is the ultimate solution to his problem? There are probably better ways to do it, but here is a simple solution I figured out.
SUM =
SUM ( Source_data[Amount] ),
ALL ( Source_data ),
Source_data[Document] = EARLIER ( Source_data[Document] )
Walking through it, The CALULATE is used to turn our row context, into a filter context. Then it manipulates that filter context so SUM “sees” only a certain set of rows.
The first manipulation is to run ALL against the table, to undo any filters applied to it. In this case, the only filter is our converted row context. (confused yet?)
The next manipulation is to use EARLIER (which is horribly named) to get the value from the earlier row context. In this case we are filtering ALL the rows, to all of them that have the same
document. Then, finally we apply the SUM, which “sees” the newly filtered rows.
Here is what we get as a result:
How do we verify that?
A fourth pain with DAX is that it’s very hard to look at intermediate stages of a process, like you can with SQL or Excel formulas, but in this case we have a way. If we convert our SUM to a
CONCATENATEX, we can output all the inputs as a comma separated list. This gives us a slightly better idea of what’s going on.
What’s the point?
My point is, that DAX, despite it’s conciseness and richness is hard to start using. Even basic tasks can require complex concepts, and that was a big frustration point for me. You can’t just google
GROUPBY and understand what’s going on.
Again, check out my presentation I did for the PASS BI virtual group. I tried to cover all the annoying parts that people new to DAX will run into. That and buy a book! you’ll need it. | {"url":"https://www.sqlgene.com/2017/07/","timestamp":"2024-11-03T20:29:18Z","content_type":"text/html","content_length":"59569","record_id":"<urn:uuid:02ea39d9-167c-4d9f-be57-d0034eb0c2f2>","cc-path":"CC-MAIN-2024-46/segments/1730477027782.40/warc/CC-MAIN-20241103181023-20241103211023-00069.warc.gz"} |
Binomial Option Pricing Model Excel with MarketXLS formula - CHIRAS Collection
Binomial Option Pricing Model Excel with MarketXLS formula
The following SAS macro is used to implement binomial trees and calculate the price of a stock, a call option, a put option, and a risk-free bond price. Each node in the option price tree is
calculated from the two nodes to the right from it (the node https://1investing.in/ one move up and the node one move down). The Binomial Options Pricing Model provides investors with a tool to help
evaluate stock options. For each period, the model simulates the options premium at two possibilities of price movement (up or down).
The model offers a calculation of what the price of an option contract could be worth today. It is also more accurate than the Black-Scholes formula, which is another popular method for pricing
options, especially for longer-term options or options that pay dividends. However, the binomial option pricing model is also more complex and time-consuming to calculate, and it may not work well
for options with multiple sources of uncertainty or complicated features.
1. Now you can price different options with the Cox-Ross-Rubinstein model – just change the inputs in the yellow cells B4-B11.
2. Call payoff is underlying price at expiration (cell L4) minus strike; put payoff is strike minus underlying price.
3. The smallest number of times the coin could land on heads so that the cumulative binomial distribution is greater than or equal to 0.4 is 5.
4. Yield can be continuous dividend yield for stock or index options, or foreign currency interest rate for currency options.
5. Where S is the underlying price tree node whose location is the same as the node in the option price tree which we are calculating.
This Excel spreadsheet implements a binomial pricing lattice to calculate the price of an option. This involves stepping back through the lattice, calculating the option price at every point.
Consider a stock (with an initial price of S0) undergoing a random walk. Over a time step Δt, the stock has a probability p of rising by a factor u, and a probability 1-p of falling in price by a
factor d. This method, first published in 1999, is more accurate than the quadratic approximation for options with small or large maturity times.
The Bjerksund & Stensland approximation was developed in 1993. For long-dated options, the Bjerksund & Stensland model is more accurate than the Barone-Adesi & Whaley method. American options do not
have closed-form pricing equations.
In reality, many more stages are usually calculated than the three illustrated above, often thousands. At each stage, the stock price moves up by a factor u or down by a factor d. Note that at the
second step, there are two possible prices, u d S0 and d u S0. If these are equal, the lattice is said to be recombining. If they are not equal, the lattice is said to be non-recombining.
Exact formulas for move sizes and probabilities differ between individual models (for details see Cox-Ross-Rubinstein, Jarrow-Rudd, Leisen-Reimer). For instance, at each step the price can either
increase by 1.8% or decrease by 1.5%. These exact move sizes are calculated from the inputs, such as interest rate and volatility. This means trinomial trees are a better description of the real-life
behavior of financial instruments.
Knowing the current underlying price (the initial node) and up and down move sizes, we can calculate the entire tree from left to right. Trinomial option pricing was proposed by Boyle (1986) and
extends the binomial method to better reflect the actual behavior of financial instruments. Both methods can be used to calculate the fair value of American and Bermudan options, and converge to the
same results at the limit.
Like the Free Spreadsheets?
5, we use Microsoft Excel programs to create large decision trees for the binomial pricing model to compute the prices of call and put options. The Black Scholes model is more reliable when it comes
to complicated options and those with lots of uncertainty. When it comes to European options without dividends, the output of the binomial model and Black Scholes model converge as the time steps
How Binomial Trees Work in Option Pricing
At the end of the year, there is a 50% probability the stock will rise to $125 and 50% probability it will drop to $90. If the stock rises to $125 the value of the option will be $25 ($125 stock
price minus $100 strike price) and if it drops to $90 the option will be worthless. This Excel spreadsheet calculates the price of a Bond option with a binomial tree. We will create both binomial
trees in Excel in the next part.
With growing number of steps, number of paths to individual nodes approaches the familiar bell curve. There are also two possible moves coming into each node from the preceding step (up from a lower
price or down from a higher price), except nodes on the edges, which have only one move coming in. Yield can be continuous dividend yield for stock or index options, or foreign currency interest rate
for currency options. If you would like access to the VBA used to generate the binomial lattice, please use the Buy Unlocked Spreadsheet option.
Appendix 23.1: SAS Programming to Implement the Binomial Option Trees
These Excel spreadsheets implement the pricing approximations described above. Any of these Excel spreadsheets can be easily adapted to calculated the implied volatility of an American option by
using Excel’s Goal Seek functionality. This article summarizes several methods for pricing American options, and provides free spreadsheets for each.
The Americal style options contracts are the ones that can be exercised on any day until the expiry. Unlike, the Black Scholes model the Binomial option pricing model excel calculates binomial tree
excel the price of the option at various periods until the expiry. Since most of the exchange-traded options are American style options, the Black Scholes model seems to have a limitation.
Creating Binomial Trees in Excel
We have completed the binomial trees – the part that is common for all the models. But our spreadsheet is not done yet, because we have used dummy values for up and down move sizes and probabilities.
Their calculation is different under different binomial models. There are a few major assumptions in a binomial option pricing model. First, there are only two possible prices, one up and one down.
Third, the interest rate is constant, and fourth, there are no taxes and transaction costs.
However, binomial methods are now outdated and, apart from being easily implemented, have no significant advantage compared to other approaches. The entire underlying price tree is centered around
the initial underlying price 100 all the way to expiration. This method gives the price of an option at multiple points in time (and not just at the expiry date, as with the standard Black-Scholes
model). Binomial trees are hence particularly useful for American options, which can be exercised at any time before the expiry date. If you have any questions or comments about this binomial option
pricing tutorial or the spreadsheet, then please let me know. Scroll down to the bottom of this article to download the spreadsheets, but read the tutorial if you want to lean the principles behind
binomial option pricing. | {"url":"https://chiras.gr/2023/07/13/binomial-option-pricing-model-excel-with-marketxls/","timestamp":"2024-11-06T07:18:28Z","content_type":"text/html","content_length":"57589","record_id":"<urn:uuid:26c52478-af14-4581-9ffe-269b35918e79>","cc-path":"CC-MAIN-2024-46/segments/1730477027910.12/warc/CC-MAIN-20241106065928-20241106095928-00593.warc.gz"} |
6 The phase plane, nullclines, stable points and separatrix. The pendulum,
6 The phase plane, nullclines, stable points and separatrix. The pendulum, Euler-Cromer eqns, SIR model of disease, bacterial growth.
6 The phase plane, nullclines, stable points and separatrix. The pendulum, Euler-Cromer eqns, SIR model of disease, bacterial growth.#
# import all python add-ons etc that will be needed later on
%matplotlib inline
import numpy as np
import matplotlib.pyplot as plt
from sympy import *
init_printing() # allows printing of SymPy results in typeset maths format
plt.rcParams.update({'font.size': 16}) # set font size for plots
6.1 Introduction#
In the study of two-dimensional linear and non-linear differential equations, the phase plane, nullclines, and fixed points are very useful tools for analysing the equations before any numerical or
algebraic calculation. The phase plane allows fixed points to be found, these are also called steady state, or equilibrium points and occur when \(dy/dt = dx/dt = 0\). Some examples are also given in
Chapter 10. Closed orbits and limit-cycles can also be observed in the phase plane and are described shortly.
Written generally, pairs of differential equations are
\[\frac{dy}{dt} = f(x, y, t) \quad\text{ and }\quad \frac{dx}{dt} = g(x, y, t)\]
where the functions \(f\) and \(g\) may contain terms in \(x,\; y\) and perhaps \(t\). For example,
\[ \frac{dy}{dt} = xy+x^2 -y^2, \qquad \frac{dy}{dt}=-\sin(x)+(y-1)^2\]
6.2 The phase plane \(^2\)#
The phase plane is the plot of \(y\) vs \(x\) and is found by calculating \(dy/dx\) by using the chain rule. In practice, this means dividing the equation for \(y\) by that for \(x\); therefore the
phase plane does not explicitly contain time. If the resulting equation can be integrated, the family of curves produced can be plotted for different values of the integration constant, which means
for different initial values of \(y\) and \(x\). If the integration cannot be done analytically, then a numerical solution of the two initial equations has to be found and, again, \(y\) plotted vs \
(x\) at various times until the entire phase plane is produced.
For example, if the equations governing the motion of a particle in a particular double-well potential, Fig. 13, are
\[ \frac{dy}{dt} =x-x^3, \qquad \frac{dx}{dt}=y\]
calculating the phase plane means integrating
\[\displaystyle dy/dx = (x - x^3)/y\]
Separating \(y\) and \(x\) gives
\[\displaystyle \int ydy=\int(x-x^3)dx\]
which integrates to
\[y=\sqrt{x^2-x^4/2 +c} \]
and this is the equation describing this phase plane. The values of integration constant \(c\) depend on the initial energy of the particle. If it has sufficient energy, then the barrier will be
surmounted and oscillation will occur between both wells. If not, the motion is restricted to one well only. The phase portrait is the collection of curves drawn on the phase plane with different
initial conditions, in this case, different values of \(c\).
The stable, steady-state, or equilibrium points are found when the rate of change is zero, and are therefore found at
\[\displaystyle dy/dt=x - x^3 =0\quad\text{and}\quad dx/dt=y=0\]
making fixed points at \(x=0,\;y=0\) and \(x = \pm 1,\; y = 0\). A more detailed analysis of fixed points in general shows that they may be stable, unstable or saddle points, see Jeffrey (1990) and/
or Strogatz (1994).
A stable steady state has the property of returning to that state after a small perturbation to it is made, and clearly, the points (\(\pm 1,\; 0\)) are of this nature, Fig. 13 as they are at the
bottom of the wells. The origin is a saddle point, and is not fully stable because moving in any direction from the origin the gradient is negative, except moving up or down the y-axis.
\(^2\) The name is historical and apparently was originally used in dynamics as the plane containing position \(x\) and momentum \(mdx/dt\) of a object as it moved under the influence of a force.
6.3 Isoclines and nullclines#
The equation produced when the rate of change is zero is sometimes called an isocline or nullcline. Just as on an incline we move up, or on a decline down, an isocline means that the gradient is
always the same and must therefore be a constant number, and a contour path is followed in the phase plane. A nullcline occurs when this constant is zero The nullclines are
\[\displaystyle x - x^3 = 0\quad\text{ and }\quad y = 0\]
which in this case are both straight lines. Isoclines are the lines when
\[x - x^3 = a\quad\text{ and }\quad y = a\]
where \(a\) is some constant. Assuming that \(y\) is plotted vertically and \(x\) horizontally, the ‘flow’ or vector showing the direction of change, is always horizontal at each point along the
nullcline, \(dy/dt = 0\), no matter what its curve is, and vertical on the nullcline \(dx/dt = 0\). The nullclines also partition the phase plane into areas where the derivatives have different sign;
exactly what these are depends upon the particular equations. Figure shows the phase portrait with isoclines at different \(c\) values and the nullclines, which are dotted. The figure-of-eight curve
is the separatrix and, in this example, is the point when the particle has just enough initial energy to cross the barrier separating the region of oscillation in one well from motion over the
barrier, and hence, motion between both wells. When the particle is placed in the bottom of either well, it has zero potential and zero kinetic energy. If it is not pushed, it will remain in this
stable state at (\(\pm 1, 0\)) which are the points on the axis marked with a red dot.
Finally, in this short introduction to the phase plane, it must be remembered that although the phase plane equation 37 does not explicitly contain time, \(x\) and \(y\) are still functions of time
and that time passing on the phase plane is not measured by equal \(x\) and \(y\) motion, but in a very non-linear manner. This can only really be observed by plotting pairs of \(x\) and \(y\)
coordinates on the phase plane at various times after solving the coupled equations.
Figure 13. Example of a phase - plane with a few contours (isoclines) making up the phase portrait. The potential energy profile has barrier with a maximum energy of zero. The motion of a particle
starting at different points is shown on the right. The \(dy/dt = 0\) nullclines are shown dotted, the other nullcline, \(dx/dt = y = 0\) is the x-axis. The arrows show the direction of motion around
the phase plane. The separatrix is the light blue ‘figure of eight’ and passes through the origin.
6.4 Non-linear equations: the pendulum#
A rigid pendulum with a heavy bob at its end can move in two ways. When the energy is small, it will oscillate about the vertical in a good approximation to simple harmonic motion, and when the
energy is large enough it will rotate continuously in the vertical plane. If the displacement from the vertical is not small, the motion is non-linear and the equation of motion has no exact
analytical solution. If it is assumed that the pivot holding the pendulum is frictionless and that no air or other resistance hinders the motion, then the equation of motion is
\[ \frac{d^2\varphi}{dt^2}+\omega^2\sin(\varphi)=0 \qquad\tag{38}\]
This equation is described as non-linear because the angle \(\varphi\) does not change linearly but as \(\sin(\varphi)\). A derivation of is given in Chapter 10.
The variable \(\varphi\) is the angle in radians away from the vertical, and \(\omega\) is an angular frequency defined as
\[\displaystyle \omega = \sqrt{g/L}\;\mathrm{s^{-1}}\]
where \(g\) is the acceleration due to gravity and \(L\) the length of the pendulum. The frequency \(\omega\) is the frequency that the pendulum has when it undergoes infinitesimally small
oscillations. The mass of the pendulum is \(m\); it is used to calculate forces but cancels out in the result.
When the angular displacement is small expanding the sine as a series and retaining the first terms as the next terms is \(\varphi^3/3!\) and will be insignificant, gives, \(\sin(\varphi) \to \varphi
\), the the pendulum’s motion is sinusoidal and that of the simple harmonic oscillator of frequency \(\omega\). The equation of motion becomes
\[ \frac{d^2\varphi}{dt^2}+\omega^2=0 \]
which has the general solution,
\[ \varphi (t)=c_1\sin(\omega t+c_2)\]
where \(c_1,c_2\) are integration constants and depend on initial conditions. These are chosen to be \(\varphi (t)=\varphi_0\) and \(d\varphi/dt=v_0\) at \(t=0\). The equations become
\[\displaystyle \varphi_0=c_1\sin(c_2)\quad\text{ and }\quad d\varphi/dt = v_0=c_1\omega\cos(c_2)\]
which after finding the constants produces
\[\varphi(t)=\varphi_0\cos(\omega t)+\frac{v_0}{\omega}\sin(\omega t) \]
Naturally, different initial conditions would lead to different \(c_{1,2}\).
Much of the dynamics of the real pendulum can be understood from the phase plane (\(\varphi,\; d\varphi/dt)\) which can be calculated easily if the equation of motion is split into two; the first
gives the angular velocity \(v\), the second the angular acceleration or rate of change of velocity,
\[\frac{d\varphi}{dt} = v, \quad \frac{dv}{dt} + \omega^2 \sin(\varphi) = 0 \qquad\tag{39}\]
The equilibrium or steady state point when the derivatives are zero is clearly \(v = 0\) and \(\omega^2 \sin(\varphi) = 0\), which will occur when \(\varphi = 0, \pm n\pi\) with \(n = 1,\, 2 \cdots\)
. The nullclines are zero because only \(v\) or \(\varphi\) occur in each equation; the isoclines are found when the derivative is a constant \(k\), whose values you can choose. In this case,
\[\displaystyle v = k\quad\text{ and }\quad\varphi = \sin^{-1}(k/\omega^2)\]
are the isoclines. The phase plane is obtained by first using the chain rule to give
\[ \frac{dv}{d\varphi} = -\frac{\omega^2}{v} \sin(\varphi)\]
and then variables \(v\) and \(\varphi\) can be separated and the equation integrated to give
\[\displaystyle v=\sqrt{2\omega^2\cos(\varphi)+2c}\]
where \(c\) is a constant of integration. This constant will be determined by the starting conditions; these are the angle that the pendulum is released from and its angular velocity at the point of
release. If the initial velocity is \(v_0\) and the release angle \(\varphi_0\) then the integration is
\[\int_{v_0}^v vdv = -\omega^2\int_{\varphi_0}^\varphi \sin(\varphi)d\varphi\]
which produces
\[v=\sqrt{v_0^2+2\omega^2[\cos(\varphi)-\cos(\varphi_0)]} \]
and this is shown in fig 14.
(i) The Separatrix#
The line crossing through \(\varphi/\pi=\pm 1\) on the abscissa, is called the separatrix; this is produced in this example when the integration constant \(c = 1\) if \(\omega = 1\). At all points
between the separatrixes, the pendulum does not complete more than one revolution, i.e. oscillates back and forth, and the motion appears as closed curves in the figure. If the pendulum starts from a
stationary position at any angle except zero, up to a fraction short of \(\pi\) radians, a position almost upside down, it will then swing, ad infinitum, to a similar position on the other side and
then back again; recall that there is no friction term in the equations. If the pendulum starts exactly upside down and also isn’t given a push, i.e. initial angular velocity is zero, then it should
remain upside down for ever in this metastable state. However, no matter what angle the pendulum is in initially, if it is given a sufficient push and acquires energy in excess of \(2mgL\), then it
can repeatedly rotate though 360\(^\mathrm{o}\). This is shown on the phase plane by the lines above the separatrix that do not cross the horizontal \(\varphi/\pi\) axis.
The direction of the motion can also be determined from the plot, starting at \(\varphi/\pi = 1/2\) or 90\(^\mathrm{o}\), and at zero initial velocity, the pendulum loses potential energy and gains
kinetic energy. The angle decreases as the pendulum moves towards its lowest point; this means that the velocity is negative and becomes increasingly so, reaching its largest negative value when the
pendulum is pointing vertically down. The motion is therefore clockwise around the closed curves as shown by the arrows on the plot. The motion continues forever, because energy is conserved in this
model of the pendulum.
Figure 14 Phase plane, angle vs velocity for the pendulum with \(\omega = 1\) and various initial conditions. The separatrix are the lines crossing at \(\varphi/\pi = \pm 1\), they separate regions
of oscillation, inside the closed curve, from complete rotation of the pendulum.
The change in angle and velocity with time can be found by numerically solving equations 39 as shown in Fig. 15 using the method outlined in Algorithm 14 with change in notation from \(x,\,y\) to \
dphidt= lambda v : v
dvdt = lambda phi : -omega0**2*sin(phi)
and changing the steps in the loop of the Euler algorithm to (commented out to show)
#v = v + h*dvdt(phi)
#phi = phi + h*dphidt(v)
#t = t + h
An exact algebraic solution is only possible when the angle is small and \(\sin(\varphi) \to \varphi\), which is the harmonic oscillator and has a frequency \(\omega\) and a period of 1/\(\omega\)
seconds. When the starting angle is not small, the angular motion is not purely sinusoidal, as may be seen in the figure, but spends longer near to the turning points at the top of the swing.
Figure 15. Angle (radians) and velocity (radians/sec) vs time of the non-linear pendulum, with \(\omega = 1,\;v_0=0\) and an initial angle of \(8\pi/9\). Notice that the velocity is zero when the
potential energy is a maximum. This is when its angle is greatest or smallest and vice versa.
6.5 Euler–Cromer equations#
Although the Euler method will work well in all our examples, it is not necessarily the best method to use for oscillating systems, such as the pendulum, because energy is not conserved that well. By
changing the algorithm as below then the error in the energy becomes proportional to \(\Delta t^3\) which is a significant improvement over \(\Delta t\) when the step size is small. The angle is
calculated as \(\mathtt{phi= phi + h*v}\) instead of \(\mathtt{phi= phi + h*dphidt}\) as in the Euler method, Algorithm 14. See Gould et al. (2007) for more details of this and related methods.
# Algorithm: Euler Cromer modifications
def EulerCromer(dphidt, dvdt, phi0, t0, maxt, omega):
v0 = 0.0
n = 1000
Eulerv = np.zeros(n,dtype=float)
Eulerphi = np.zeros(n,dtype=float)
dtime = np.zeros(n,dtype=float)
h = (maxt-t0)/n # time step
v = v0 # initial values
phi = phi0
t = t0
Eulerv[0] = v0
Eulerphi[0] = phi0
dtime[0] = t0
for i in range(1,n):
v = v + h*dvdt(phi,v,t)
phi = phi + h*v
t = t + h
Eulerv[i] = v # save values
Eulerphi[i] = phi
dtime[i] = t
return Eulerv,Eulerphi,dtime
7 The SIR model describes the spread of diseases#
A very interesting, and relatively straightforward example of coupled equations is the spread of an infectious disease, because, besides being intrinsically interesting, especially during the Covid19
epidemic, it allows a clear illustration of a number of features such as the phase plane and nullclines.
An epidemic is defined as the number of infected persons, increasing with time to a number above those initially infected. Kermack & McKendrick (1927), were the first to describe a realistic disease
model, which they used to study the spread of a plague on the island of Bombay in 1905/6. In the SIR model, one or more infected persons are introduced into a community where all are equally
susceptible to the disease. The model assumes, first, that the disease spreads by contact one to another; each person runs the course of the disease and then cannot be re-infected; and, secondly,
that the duration of the infection is short compared to an individual’s lifetime, so that the total number of people is constant. Finally, the number of individuals is fixed once the infection has
begun; therefore this model only describes infection in a closed community and is called a compartmentalised model. This is called the S-I-R model, because individuals are either susceptible (S),
infected (I), or removed (R). The scheme is
\[ S+I \overset{k_2}\longrightarrow 2I ; \qquad I \overset{k_1} \longrightarrow R \qquad\tag{40}\]
and the aim is to calculate how R, S, and I change with time. In doing so we shall set up a set of rate equations and suppose that the numbers of individuals involved is sufficiently large that the
integration is valid. In the case of Covid-19 the numbers are so large that this should not present a problem.
The first step describes the transmission of the infection, and the second, the recovery from infection, hence, the number infected I must reach zero at long times as the infection ends. The
susceptible persons become infected by reacting with someone who is already infected with a rate constant \(k_2\). In chemical terms, the rate of such a second-order reaction is \(k_2[S][I]\),
supposing that \([S]\) and \([I]\) are concentrations. The second step, infected to removed, has a rate constant \(k_1\), the reciprocal of which is the average time that an individual once infected
takes to move into the removed class; the rate for this is \(k_1[I]\). Out of the constant total number of individuals, \(N = S + I + R\); the number \(R_0\) before the infection starts are those
that are immune and clearly at least one infected person has initially to be present. The second equation shows that given time, all individuals will end up in the removed class R, and play no
further part in the infection, being immune, isolated, or dead. In this model, the epidemic is assumed to run its course without the intervention of medication, which, given at random times to
different individuals, would prematurely cause its end.
If S, I, and R were chemical species, the scheme above would represent a quadratic autocatalytic reaction where R is the product that takes no further part in the reaction. To start the reaction,
some initial amount of species I has to be present.
7.1 Rate equations#
The rate equations for scheme 40 are
\[\begin{split}\qquad\qquad \displaystyle \frac{dS}{dt} &= -k_2SI \\ \displaystyle \frac{dI}{dt} &= +k_2SI - k_1I\\ \displaystyle \frac{dR}{dt} &= +k_1I \\ N &= I + S + R \qquad\qquad\qquad\qquad\
where \(S\) and \(I\) represent the number of individuals (or the concentration of chemical species in an autocatalytic reaction). The initial number infected is \(I_0\), those susceptible \(S_0\),
and removed \(R_0\). The total number of individuals is a constant \(N\), and, because of this, the last differential equation is not needed because \(R\) can be calculated by subtracting from \(N\)
the amount of \(S\) and \(I\) at any time. The rate constants are \(k_2\), the spreading rate constant and \(k_1\) the removal rate constant, and have units of number\(^{-1}\) time\(^{-1}\) and
number time\(^{-1}\) respectively. By writing down rate equations, it is implicitly assumed that the number of individuals present is large and can be a continuous variable, not an integer, as is
really the case. The model described so far assumes that all infected persons recover but instead many may die as we have seen with Covid19. To model this an extra rate constant can be added to \(k_1
\to k_1+k_D\) and then the term \(dD/dt=k_DI\) included. The fraction that die is \(k_1/(k_1+k_D)\), a number which will normally be known. Monte Carlo methods to integrate and simulate these
equations without using calculus are shown in Chapter 12 in Q9 and Q16 and their answers.
Before the equations are numerically integrated, a complete analytical solution not being possible, some analysis of the problem can still be carried out. The actual values of the constants are
important if a real disease is to be modelled, and before trying to fit the data, it is necessary to know what range of parameters will produce an epidemic and what the expected populations will look
Intuitively, scheme 40 suggests that the number infected I, which is initially small (for instance, one person), increases rapidly, passes through a maximum, then slowly decays away. However, this
will occur only if \(k_2S_0 \gt k_1\) because when S is large, I is initially formed more rapidly than it is consumed. If the opposite is true, then I is consumed more rapidly and its population
cannot become large and no epidemic occurs. To be quantitative, let \(R_R = k_2S_0/k_1\) be defined as the reproductive ratio,\(^\dagger\) which is the number of secondary infections caused by one
infected person if all the population is equally susceptible. An epidemic must ensue if the reproductive ratio is greater than one, because more individuals will become infected with time. This can
also be appreciated by examining the rate of change of I in the second of equations 41 at
\[\displaystyle t = 0;\; dI/dt\big|_{t = 0} = I_0(k_2S_0 − k_1)\quad\text{ if }\quad k_2S_0 \gt k_1\]
then \(dS/dt \gt 0\) and an epidemic will occur.
Typical values for the reproductive ratio are smallpox = 4; mumps = 5; German measles (rubella) = 6; measles = 12; malaria \(\approx\)100, (see Britton 2003), Covid-19 \(\gt 3 \lt 9\) (Wikipedia).
( \(^\dagger\) Many texts call the reproductive ratio \(R_0\), which is unfortunately confusing with \(R_0\), the initial number in the removed class.)
7.2 The SIR phase plane#
A graph of \(I\) vs \(S\) is the phase plane. The phase plane shows how the number of infectives \(I\) and susceptibles \(S\) change with time, even though time is only implicit on the graph. The
relationship between I and S is found by using the chain rule
and then integrating:
Integrating with limits \(I_0,\;S_0\) gives
\[\displaystyle \int_{I_0}^IdI=\int_{S_0}^S \left(\frac{k_1}{k_2S}-1\right)dS\]
which produces
\[I=\frac{k_1}{k_2}\ln\left(\frac{S}{S_0} \right)-S+N\]
where the initial values have been substituted with \(I_0 + S_0 = N\); at \(t = 0,\; R = 0\). Next, dividing by \(N\) to make the calculation independent of the number of individuals produces
\[I_N=\frac{k_1}{k_2N}\ln\left(\frac{S_NN}{S_0} \right)-S_N+1 \qquad\tag{42}\]
where the notation is \(I_N = I/N\) and similarly for \(S_N\). The graph of \(I_N\) vs \(S_N\) is shown in Fig. 16 at different values of \(S_0/N\), which is the fraction initially susceptible. Time
does not explicitly appear in this equation, but this does not mean that the curves are time independent; far from it, because S and I both depend on time. The curves, equation 42, must start at the
line \(S_N +I_N =1\) , or \(S+I=N\), which is the diagonal line in Fig.16, because no \(R\) (removed class) individuals are present initially, and must move to the left as time progresses. At very
long times the fraction infected must become zero.
Figure 16 Phase plot of equation 42. Different fractions of initially susceptible individuals \(I_N\) are shown calculated with \(S_N^{max} = 1/3\). An epidemic occurs when a curve starts to the
right of \(S_N^{max}\). The arrow shows the direction of change with time and are horizontal on the (vertical) nullcline, \(S = k_1/k_2\), and vertical on the \(I = 0\) nullcline or horizontal axis.
The definition of an epidemic is that the number of individuals infected increases above those infected initially. In Fig. 16, the initial number infected is found where a curve touches the diagonal
line, this is 0.25 with \(S_0/N = 0.75\), and \(I_N\) increases to \(\approx 0.4\) at its maximum and therefore, an epidemic may occur. Starting at \(S_0/N = 0.25\), the number infected decreases
continuously and therefore an epidemic cannot occur. This simple approach indicates the importance of immunization and vaccination.
Immunizing or vaccinating a population reduces those susceptible, reducing \(S_0\) and the reproductive ratio \(R_R\), and so making an epidemic less possible. To the right of \(S_N^{max}\), Fig.16,
an epidemic occurs, although it may not be severe if the initial value of \(S_0/N\) is close to the maximum; to its left the infection dies out. The turnover from epidemic to no epidemic is the point
where \(S_N^{max}\) touches the diagonal. In the figure this occurs at \(I_N^{max} = 1 - 1/3 = 0.66\), meaning that \(66\)% immunization is needed to prevent an epidemic, which is a low value. With
an infectious disease such as mumps or German measles, this value has to be \(\approx\) 0.85, meaning that 85% of the population has to be immunized to prevent an epidemic. Notice that not everyone
needs to be immunized to prevent an epidemic; this is called herd immunity. A few individuals will by chance, never meet an infected person. An immunization/vaccination level of \(85\)% may be
difficult to achieve in a population by voluntary mass vaccination. Should the level of immunization fall by only a small amount, the threshold at \(S_N^{max}\) may be crossed and an epidemic could
occur. The number of individuals being immunized can suddenly fall, as happened in the UK in the late 1990’s and early in this century, due to poorly researched and inflammatory news media stories
about the MMR vaccine for children. Some parents were reluctant to have their children vaccinated even though the risks of damage to health and even death were far greater than receiving the vaccine
itself. Similar concerns have prevented many people from becoming vaccinated for Covid19 even though world-wide multiple millions of doses have been administered.
7.3 Steady states, isoclines, and nullclines#
In a rate equation, a steady state is produced when the rate of change is zero. In the SIR model this means
\[\displaystyle \frac{dS}{dt} = 0,\qquad \frac{dI}{dt} = 0\]
When molecules, or species in general, interact more than one steady state can be present, and not all of these are necessarily stable. The nullclines on the phase plane of the SIR model are
particularly simple and are \(I = 0\), or along the S-axis, and \(S = k_1/k_2\), which is the vertical line at \(S_N^{max}\) and divides the region where the infected population increases from that
where it decreases. The nullclines divide the phase plane into four areas, two areas are below the S-axis in this case, and, as negative number of individuals do not make any sense, only two of the
four regions have any meaning. Assuming that I is plotted vertically and R horizontally, the ‘flow’ or vector, arrow Fig.16, showing the direction of change is always vertical at any point of the \(I
\) nullcline, (horizontal axis), no matter what its curve is, and horizontal on the \(S\) nullcline when \(dS/dt = 0\). A steady state point is found where the nullclines meet, in the SIR model this
is in the \(S\)-axis at the point \([k_1/k_2, 0]\) which is the foot of the vertical line \(S_N^{max}\).
7.4 Threshold for an epidemic and maximum and total number infected#
The maximum fraction of infected individuals is found when \(dI_N/dS_N = 0\), and this occurs at the constant value \(\displaystyle S_N=\frac{k_1}{k_2N}\) for any fixed \(N\). When \(dI_N/dS_N=0\) is
reached, the infection has peaked and must start to decrease. The maximum fraction infected at any one time, is from 42
\[I_N^{max} = 1+\frac{k_1}{k_2(S_0+I_0)}\left(\ln\left( \frac{k_1}{k_2S_0} \right) -1 \right) \qquad\tag{43}\]
and, when multiplied by the total number \(I_0 + S_0\), gives the maximum number of hospital beds necessary to treat the infection. The maximum \(I_N^{max}\) may occur mathematically to the right of
the diagonal line, Fig. 16, but clearly this is not physically possible, because the maximum value \(I_N\) can ever take is subject to the condition \(S_N^{max} + I_N^{max} \le 1\) and this occurs
when \(R_R \ge 1\).
When the number (or fraction) of infected individuals is zero, there are still some who remain susceptible, see Fig. 16, who did not catch the disease even in an epidemic. This fraction is in the
range \(0.02 \to 0.08\) in the figure, and is the extent of herd immunity. To be more quantitative, equation 42 describes \(I\) vs \(R\), and when \(t \to \infty\) then \(I\) is zero, giving
\[0 = \frac{k_1}{k_2N}\ln\left( \frac{S_N^\infty N}{S_0} \right)-S_N^{\infty} +1 \qquad\tag{44}\]
which is transcendental, and has to be solved numerically for \(S_N^\infty\), the fractional amount of \(S\) remaining at \(t \to \infty\). The Newton-Raphson method (Chapter 3.10) could be used to
solve the equation. However, for a strong epidemic the fractional amount of \(S\) left at the end is very small; \(S_N^\infty \ll 1\) hence
\[\frac{k_1}{k_2N}\ln\left( \frac{S_N^\infty N}{S_0} \right) +1=0 \qquad\tag{45}\]
When re-arranged,
\[S_N^\infty =\frac{S_0}{N}e^{-k_2N/k_1},\qquad \text{or}\qquad S_N^\infty =\frac{S_0}{I_0/S_0+1}e^{-k_2S_0/k_1(1+I_0/S_0)} \qquad\tag{46}\]
When \(I_0/S_0 \ll 1\), for example, if only one person is infected initially, then \(\displaystyle S_N^\infty \approx e^{-k_2S_0/k_1}\). As a check, using the lowest curve in Fig. 16, which has been
calculated with the ratio \(k_2N/k_1 = 3, S_0 = N - I_0\) and \(I_0/S_0 = 1/1000\), the fractional amount of \(S\) remaining at the end of the epidemic, calculated using the approximate formula, is \
(S_N^\infty \approx 0.0498\) or 4.98% of the population were never infected. This is close to the exact value of 5.94%.
Finally, starting with eqn. 45 the total number infected is approximately
\[I_{tot} \approx N-S_0e^{-k_2N/k_1} \qquad\tag{45}\]
or \(N(1 - 0.0498)\) and this is the size of the infection, and in effect defines the total number of hospital beds needed over the course of the epidemic.
7.5 Sensitivity to rate constant \(k_2\)#
To understand some of the equations above, such as the maximum fraction infected, \(I_N^{max}\) it is easier to draw some graphs. Suppose that there are ten thousand susceptible persons, and that the
disease starts to spread after one person is infected. What is the effect of the different rate constants? The rate of infection \(k_2\) here is key, i.e. the term \(k_2SI\) (eqn. 41) has a dramatic
effect. If \(k_2\) is small, \(10^{-5}\), top left, (see the graphs below), the overall the epidemic does not progress even up to \(1000\) days. However, with \(k_2=2\cdot 10^{-5}\) (top right) the
epidemic starts and \(\approx 1500\) persons are infected at the peak. Increasing \(k_2\) again to \(5\cdot 10^{-5}\) \(\approx 4800\) are infected. This rapid increase is rather dramatic and shows
the effect of feedback, i.e. \( S+I \overset{k_2}\longrightarrow 2I \); Increasing \(k_2\) above \(10^{-4}\) has a smaller effect than when \(k_2=10^{-5}\) simply because so many are already
infected. This behaviour shows how important the reproductive ratio is because this is proportional to \(k_2\).
Even with this simple model we can understand why wearing face masks is important because this effectively reduces \(k_2\). Isolating individuals, or being vaccinated, reduces the spread of disease
also because the number of those susceptible is reduced and therefore so is the product \(k_2SI\) in eqn. 40.
Figure 16a. Populations of susceptibles S, (red line), infected I, (blue line) and recovered R, (green line) vs. time with different \(k_2\) values using \(k_1 =0.1\). The peaks of the infected
curves are found using eqn. 43, \(I_N^{max}N\).
7.6 Calculating the time profile of an epidemic#
As an illustration of using the SIR model, suppose that you are presented with this specific problem: The following data for the incidence of influenza was recorded at a boys’ boarding school.
Starting on day zero, the number of boys infected each day was
\[\displaystyle \qquad\qquad 1, 3, 7, 25, 72, 222, 282, 256, 233, 189, 123, 70, 25, 11, 4\]
One infected boy started the flu epidemic, and \(763\) boys were resident. This example is given by Murray (2002, chapter 10, p. 326). The SIR model can be used to estimate the rate constants
describing the data, the maximum number infected at any time, and the total number of boys that have been infected at the end of the epidemic. The strategy is to work out what is already known from
the data.
First, the timescale is in days; the number of boys susceptible is \(763\), making \(S_0 = 763 - 1\), assuming that only one boy was initially infected. If the infectious period is about 2.5 days,
this means that, by definition, \(k_1 = 1/2.5\), which can be used as a starting value leaving only \(k_2\) to be estimated. We know that \(R_R = k_2S_0/k_1\) has to be greater than 1 and if all the
boys are susceptible except one,then \(k_2 \times 762 \times 2.5 \gt 1\) which makes \(k_2 \gt 5 \cdot 10^{-4}\); the maximum value of \(R_R\),is approximately 20 for common infections, making \(k_2
\lt 0.01\) and this should give a range of rate constants to start the calculation.
Only two quantities need to be calculated; the third, \(R\), is evaluated via \(I + S + R = N\). The Euler method code to integrate the differential equations is outlined in Algorithm 14 or 15. Note,
that in the calculation, instead of using I we use In. The calculation is
# Algorithm: SIR model of disease
def EulerSIRint(S0,In0, k1,k2):
h = (maxt - t0)/num
EulerS = np.zeros(Np,dtype=float)
EulerIn = np.zeros(Np,dtype=float)
dtime = np.zeros(Np,dtype=float)
EulerS[0] = S0
EulerIn[0] = In0
dtime[0] = t0
S = S0
In = In0
t = 0
for i in range(1,Np):
S = S + h*dSdt(S,In)
In= In + h*dIndt(S,In)
EulerS[i] = S
EulerIn[i] = In
dtime[i] = t
t = t + h
return dtime, EulerS, EulerIn
k2 = 0.00218 # initial k's
k1 = 0.452
num = 763 # number of individuals
dSdt = lambda S, In : -k2*S*In # eqns 41
dIndt = lambda S, In : k2*S*In-k1*In
t0 = 0.0
maxt = 25
S0 = num-1
In0 = 1
R0 = 0.0
Np = 1000 # number of points for integration
dtime, S, In = EulerSIRint(S0,In0,k1,k2)
#plt.plot(dtime, S) # remove first # to plot
#plt.plot(dtime, In)
The result of numerical integration is shown in Fig. 17, with \(k_2 = 0.0022\), day\(^{-1}\), and \(k_1 = 0.451\) day\(^{-1}\); the data were fitted as outlined in 18. The number of susceptible
individuals initially changes slowly because the product SI in the first rate equation 41 is small, \(I\) being small. As \(I\) increases, \(S\) also decreases, but, their product is larger (\(1 \
times 763\) is smaller than \(2 \times 762\) and so forth) and therefore the population of \(S\) starts to decrease rapidly as that of \(I\) increases; see the first of equations 41. The population
of \(I\) goes through a maximum because in the second equation, the term \(k_2I\) eventually starts to dominate and removes \(I\).
The maximum number infected at any time is calculated with equation 43; substituting in the numbers gives 286 boys infected, which is seen on the graph where it peaks between days \(6\) and \(7\).
The number remaining susceptible is not zero at the end of the calculation indicating that not all the boys became infected even though they were all susceptible. The total number of boys who were
not infected, using the approximation in equation 45 is 19, making \(744\) who contracted the disease. The numerical calculation indicates that \(24\) boys were not infected, showing that the
approximation, equation 45, is quite good.
Figure 17. SIR data calculated for boys infected with influenza with \(k_2 = 0.0022\), day\(^{-1}\) and \(k_1 = 0.451\) day\(^{-1}\) with \(763\) boys one of whom was initially infected. The scheme
is \(\displaystyle S+I \overset{k_2}\longrightarrow 2I ; \; I \overset{k_1} \longrightarrow R \).
After 2019 there is no shortage of data from covid-19 to illustrate the SIR type of epidemic, however, the real data only partly follows this scheme because of medical intervention, i.e. learning how
best to deal with very ill patients means that the rate constants vary with time which complicates the calculation, as do the use of face masks and of course the availability, or otherwise, of
7.7 Bacterial populations calculated via Chemical Kinetics#
Foods provide an environment for microbes to survive and multiply because they are rich in nutrients. While some microbes are harmless others such as yeasts and molds spoil foods. Bacteria, such as
the pathogens staphylococcus aureus and E. coli, also produce enterotoxins (protein toxins) which target the gastrointestinal tract and cause diarrhea and food poisoning.
The life cycle of bacteria has four stages,
(1) the lag (induction) phase when populations are small,
(2) exponential growth,
(3) a maximum population during the stationary phase, and
(4) death when the population declines.
and the population therefore rises rapidly then decreases slowly not unlike the infected population in fig 17 above. The bacterial population is often inferred from the optical density of a sample
calibrated against known standards.
Bacteria multiply by dividing so that a naive model of their increase in number follows the series \(1, 2, 4, 8, \cdots 2^n\) which would naturally lead to an infinite population. The Malthusian idea
is that growth occurs exponentially,
\[M=M_0e^{kt},\qquad \text{Malthusian}\]
with rate constant, \(k\) which is the difference between birth and death rate constants. This model is a good description when the population is relatively small but still predicts an infinite
population eventually. The Gompertz model (a variation on the Logistic equation) has been used as a way of predicting a bacterial population, this equation has the basic form
\[\displaystyle \ln\left(\frac{M}{M_0}\right) = -e^{-kt},\qquad\text{Gompertz}\]
and a plot of \(M\) vs. \(t\) looks like a sigmoidal curve becoming constant at long times. This equation, while an improvement, only approximately fits the data because microbial populations can and
will eventually die out.
Using a chemical kinetics based model of growth and death provides a rationale for describing bacterial populations. The bacterial are treated as if they were molecules that can be described by rate
equations. The model used by Taub et al. (J. Food. Sci. p2350, v68, 2003) allows bacteria to grow by division and die away naturally as well as by reaction with an antagonistic molecule produced by
the bacteria itself. This latter species is an essential feature of the model. We let bacteria, labelled M, divide into two others which are labelled A to distinguish them and additionally an
antagonist X is produced. The fact that \(A\to 2A\) provides positive feedback, or autocatalysis, so that the amount of A increases very rapidly. The species X interacts with bacteria A and cause
them to die, forming species D. The bacteria also die naturally. The scheme is
\[\begin{split}\displaystyle & M \overset{k_1}\longrightarrow A\\ & A \overset{k_2}\longrightarrow 2A+X\\& A+X \overset{k_3}\longrightarrow D\\& A \overset{k_4}\longrightarrow D\\& M \overset{k_5}\
longrightarrow D \end{split}\]
The first reaction is the lag phase, the second exponential growth and the and the third causes a limit to the populations and starts the death phase. The rate equations are
\[\begin{split}\displaystyle \frac{dM}{dt} &= -(k_1+k_5)M \\ \frac{dA}{dt} &= +k_1M +k_2A -(k_3X +k_4)A\\ \frac{dX}{dt} &= k_2A -k_3XA\end{split}\]
and the initial conditions are at time zero M is present as \(M_0\) bacteria, and \(A = X = 0\). Before calculating the populations quite a lot can be understood by examining these equations and to
do this we shall insist that the bacteria are growing normally, i.e. the maximum population is orders of magnitude greater than the initial population. From the equations we can infer that,
(i) Species M decays exponentially with rate constant \(k_1+k_5\).
(ii) Maximum X happens when \(X\) reaches steady-state, \(dX/dt = 0\) and when integrated produces a constant. The maximum possible value is \(X_{max}=k_2/k_3\) and this ratio has to be much greater
than \(1\) as the bacterial population is growing.
(iii) When species X is small and \(k_1M\) is also small, species A increases exponentially with a rate constant \(k_2-k_4\) assuming that \(k_2 \gt k_4\).
(iv) At long times after the maximum bacterial population is passed the bacteria decay exponentially with rate constant \(k_4\). At these times \(k_1M\) is very small vs. population of A, \(k_2-k_3X
\sim 0\) (see (iii) above), thus \(dA/dt \sim -k_4 A\) which integrates to an exponential decay.
(v) The rate constant \(k_3\) has to be small compared to the others. Since \(k_3XA\) is the product of two potentially large numbers A and X then \(k_3\) has to be very small to be comparable to
other rate constants. If \(10^8\) bacteria are to be produced then \(1/k_3 \sim 10^8\)
The Euler method can be used to integrate the rate equations and using \(1000\) time steps is plenty. The code for the SIR model above can be changed to do this calculation. The time-scale needed
from experience of food going bad is only a few days, thus we guess rate constants in terms of time units in days to illustrate the behaviour of the population. Some initial numbers are tried just to
get going, even if fitting to a data set using a non-linear least squares method. The graph below shows the populations of M, A and X on a linear and log scale vs. time. The initial population \(M_0
= 10000\), the rate constants are \(k_1 = 1, k_2 = 4, k_3 = 1\cdot 10^{-8}, k_4 = 0.5, k_5 = 0.01\). You can see that with this set of rate constants that the lag phase is \(\approx 3\) days, (fig
17a left), the exponential rise is very rapid about a day, and large \(\sim 10^4\) times increase, the stationary phase short, \(\approx 1\) day and the death phase long, several days. This model
therefore shows all the features of the growth and death of a bacterial population.
The fall in the population after reaching a maximum is due to \(k_4A\) becoming greater than \(k_2-k_3X\). As species A decreases the reaction \(A+X\to D\) is slowed but X is still being formed from
\(A\to 2A+X\), so this is slowed also as A decreases under the influence of \(k_4\). The result is that X becomes constant at long times since the amount formed during the exponential growth phase
remains because it is no longer being formed or removed at any appreciable rate as can be seen in fig 17a.
Figure 17a. The same calculated profile of a bacterial population on a linear scale (A) and log scale (B) and the A-X phase plane (C). The rate constants used were \(k_1 = 1, k_2 = 4, k_3 = 1\cdot10^
{-8}, k_4 = 0.5, k_5 = 0.01\). The equations in (B) show the limits as described in the text applicable when \(k_2\gt k_4\). In figure (A), M is multiplied by \(10^4\) so that it can be seen on the
same plot as A and X. The vertical lines on the phase plane (C) show the maximum possible amount of X which is \(k_2/k_3\) and at \(X= (k_2-k_4)/k_3\) the maximum A occurs. This is given by \(\
displaystyle A_{max}=\frac{k_4}{k_3}\left(\ln\left(\frac{k_4}{k_2}\right) -1\right)+\frac{k_2}{k_3}\).
The phase plane \(dA/dX\) can be found using
\[\displaystyle \frac{dA}{dt}=\frac{dA}{dX}\frac{dX}{dt}\]
which gives, after ignoring M as its value is tiny compared to A or X most of the time,
\[\displaystyle \frac{dA}{dX}=\frac{k_4}{k_3X-k_2} + 1\]
Integrating to find A as a function of X, gives
\[\displaystyle A= \frac{k_4}{k_3}\ln(k_3X-k_2) + X + C\]
where \(C\) is a constant. The initial conditions are that at \(t=0, A=X=0\) making \(C=k_4\ln(-k_2)/k_3\) therefore
\[\displaystyle A= \frac{k_4}{k_3}\ln\left(1-\frac{k_3X}{k_2}\right) + X\]
The maximum values of X and A are are
\[\displaystyle X_{max}=\frac{k_2-k_4}{k_3},\qquad A_{max}=\frac{k_4}{k_3}\ln\left(\frac{k_4}{k_2}\right) + X_{max}\]
The idea of the phase plane means that if the rate constants are known, even if only approximately, then the maximum populations can easily be calculated without having to integrate the rate | {"url":"https://applying-maths-book.com/chapter-11/num-methods-D.html","timestamp":"2024-11-02T14:12:48Z","content_type":"text/html","content_length":"134398","record_id":"<urn:uuid:485e082c-b38e-4b2c-857a-9ee5d506f49c>","cc-path":"CC-MAIN-2024-46/segments/1730477027714.37/warc/CC-MAIN-20241102133748-20241102163748-00454.warc.gz"} |
Get Workbook Path Only - Free Excel Tutorial
This post will guide you how to get the current workbook path in Excel. How do I insert workbook path only into a cell with a formula in Excel.
Get workbook path with Document Location
You can get the workbook path from the document location feature, and copy the location to a cell. Do the following steps:
#1 go to File tab, and click Options menu from the popup menu list. And the Excel Options dialog will open.
#2 click Quick Access Toolbar option, and select All commands from the drop down list of the Choose commands from. And then select Document Location value, click Add button to add it into Quick
Access Toolbar.
#3 click OK button. The workbook path is displayed in the Quick Access toolbar. Press Ctrl +C to copy it.
Get workbook path with Formula
You can also use a formula based on the LEFT function, the Cell function, and the FIND function to get workbook path only in Excel. Just like this:
Type this formula into a cell and then press Enter key.
Let’s see how this formula works:
The Cell function will be used to get the full name and path of the workbook file.
The Find function will return the location number of the first left square bracket.
The left function will extract the workbook path based on the number returned by the Find function.
Related Functions
□ Excel Find function
The Excel FIND function returns the position of the first text string (substring) from the first character of the second text string.The FIND function is a build-in function in Microsoft
Excel and it is categorized as a Text Function.The syntax of the FIND function is as below:= FIND (find_text, within_text,[start_num])…
□ Excel LEFT function
The Excel LEFT function returns a substring (a specified number of the characters) from a text string, starting from the leftmost character.The LEFT function is a build-in function in
Microsoft Excel and it is categorized as a Text Function.The syntax of the LEFT function is as below:= LEFT(text,[num_chars])…
□ Excel CELL function
The Excel CELL function returns information about the formatting, location, size, or contents of a cell.The syntax of the CELL function is as below:= CELL (info_type,[reference])… | {"url":"https://www.excelhow.net/get-workbook-path-only.html","timestamp":"2024-11-04T07:36:13Z","content_type":"text/html","content_length":"86673","record_id":"<urn:uuid:1e9fd109-c243-4ea8-9475-983aade4646d>","cc-path":"CC-MAIN-2024-46/segments/1730477027819.53/warc/CC-MAIN-20241104065437-20241104095437-00037.warc.gz"} |
Useful non-UCLA Stata programs
These are useful Stata programs from around the world We assume that you are running Stata and have the most up to date version of Stata (i.e., that you have run the update all command recently) A
number of these programs are from the Stata Technical Bulletin (STB) courtesy of, and Copyright, Stata Corporation
ANOVA tools
Plots of ANOVA fit, including interaction plots (SJ-4-4/gr0009) search anovaplot
Pairwise comparisons of means, including the Tukey wsd (STB47/sg101) net from http://wwwstatacom/stb/stb47 net install sg101
Correlation and regression tools
Box-tidwell and exponential regression models search boxtid
index plots after estimation ssc install indexplot
Multiple regression with missing observations for some variables view net install regmsng, from(“http://digital.cgdev.org/doc/stata/MO/Misc”)
Logistic, poisson, and negative binomial regression tools
module to estimate generalized ordered logit models (SJ-6-1/st00-7) ssc install gologit2
Mean score method for missing covariate data in logistic regression (STB58/sg156) net from from http://wwwstatacom/stb/stb58 net install sg156
module to calculate out-of-sample predictions for regression, logistic (STB58/sg157) ssc install predcalc
commands for the post-estimation interpretation of regression models
net from http://wwwindianaedu/~jslsoc/stata
Survey data analysis
Correlation tables for survey data ssc install corr_svy
Survey sampling weights: adjustment and replicate weight creation ssc install survwgt
Predicted means or proportions for nominal predictors for survey data ssc install svypxcat
Predicted means or proportions for a continuous predictor for survey data ssc install svypxcon
Data management tools
A labels editor for Windows and Macintosh (STB51/dm561) net from http://wwwstatacom/stb/stb43 net install dm56
How can I list observations in blocks? (STB50/dm68) net from http://wwwstatacom/stb/stb50 net install dm68
Modules for managing value and variable labels ssc install labutil
Other data analysis tools
Software to Interpret and Present Statistical Results (see documentation) net from http://gkingharvardedu/clarify/ net install clarify
Make regression tables that look like those in journal articles (SJ-5-3/st0085) search estout
Estimates generalized linear latent and mixed models ssc install gllamm
Multiple imputation for missing data (SJ-5-4/st0067_2) search ice
Inter-quartile range, including outliers search iqr
Displays the missing value patterns search mvpatterns
Can I quickly see how many missing/nonmissing values a variable has? (SJ-5-4/dm67_3) ssc install nmissing
Can I make regression tables that look like those in journal articles? (STB59/sg973) ssc install outreg
Runs MLwiN via Stata search runmlwin http://www.bristol.ac.uk/cmm/software/runmlwin/
Cross-tabulates three variables and displays any combination of cell frequencies, cell percents, row percents and column percents ssc install tab3way
How can I get descriptive statistics and the five number summary on one line? (STB51/sg671) net from http://wwwstatacom/stb/stb51 net install sg67_1 | {"url":"https://stats.oarc.ucla.edu/stata/ado/world/","timestamp":"2024-11-03T03:01:05Z","content_type":"text/html","content_length":"40513","record_id":"<urn:uuid:5e89ce0d-a229-46a0-a95b-d7d9068ec109>","cc-path":"CC-MAIN-2024-46/segments/1730477027770.74/warc/CC-MAIN-20241103022018-20241103052018-00218.warc.gz"} |
Linear Algebra in Computer Science - Unlocking the Power of Algorithms
Linear algebra is an essential tool in computer science, facilitating the development and understanding of several cutting-edge technologies. It provides a framework for managing and manipulating
multi-dimensional data structures, which is pivotal in areas like graphics, machine learning, and big data analysis.
The discipline’s basic concepts like vectors, matrices, and tensor operations form the bedrock upon which complex algorithms and data processing techniques are built. In my journey through computer
science, I’ve found these concepts are not only fundamental to comprehending how problems are structured but are also crucial in formulating efficient solutions.
Applications of linear algebra in computer science are vast and varied, ranging from internet search algorithms, which rely on vector spaces for ranking pages, to computer vision, where matrix
operations are key to image recognition.
Learning about how these mathematical structures enable us to analyze large datasets has changed my perspective on the role of mathematics in technology. Recognizing patterns, making predictions, and
even simulating entire worlds in video games are made possible through linear algebra, revealing its profound impact on the industry.
My exploration into this field has unequivocally shown me that whether we are lurking behind the scenes of a Google search or unlocking new capabilities in robotics, the traces of linear algebra are
there, proving that this area of mathematics is, quite literally, shaping our digital world.
How Linear Algebra is Used in Computer Science
In my journey through computer science, I’ve found that linear algebra is not just a branch of mathematics, but a powerful tool that underpins various subfields within this discipline. It fascinates
me how linear algebra provides the foundation for dealing with linear equations and matrices, which are indispensable in computer algorithms.
When I look at optimization problems, especially in machine learning, linear algebra is the key to finding the best parameters that minimize or maximize a certain function. For example, in linear
regression, the goal is to find the best-fit line through a set of points.
This involves solving for coefficients that minimize the difference between the predicted values and the actual data points, a process termed least squares.
Deep learning, a subset of machine learning, relies on linear algebra to manage and manipulate high-dimensional data. Here, neural networks use tensor operations which are generalizations of matrices
to higher dimensions. These tensor computations are essential when performing tasks like image and speech recognition.
In the realm of computer graphics and geometry, linear algebra is the backbone for transformations, projection, and manipulation in three-dimensional space. This includes operations like rotation,
scaling, and translation, which are fundamental to rendering images and animations.
The more I explore data science, the clearer it becomes that linear algebra also plays an integral role. Techniques like singular value decomposition (SVD) and principal component analysis (PCA)
benefit from linear algebra to perform dimensionality reduction.
By using eigenvalues and eigenvectors, these methods help identify the most relevant features in large datasets, which improve the performance of classification and recommendation systems.
Linear algebra even steps into the quantum realm. Quantum computing utilizes linear algebra for state representation and the operations that change these states. Gates in quantum computing are
represented by unitary matrices—a concept I find fascinating!
Let me give you a glimpse into the typical linear algebra entities and their applications:
Entity Application in Computer Science
Matrices Represent and solve systems of linear equations, image transformations
Vector Spaces Describe directions and shapes, subspaces in graphics
Determinants Calculate the area, volume, and invertibility of matrices
Inverse Matrices Solve linear systems, perform matrix operations
It’s clear to me that linear algebra isn’t solely about crunching numbers; it’s a diverse toolset that enables advancements across the whole spectrum of computer science.
In my journey through computer science, I’ve found that linear algebra is not just a subject studied in the classroom; it’s a powerful tool that underpins various domains within the field. For
instance, machine learning algorithms often hinge on matrix operations and vector spaces.
The optimization of these algorithms requires a clear understanding of concepts like eigenvalues ($\lambda$) and eigenvectors ($\vec{v}$ ).
Furthermore, my experience with computer graphics has shown me the importance of linear transformations and matrices in rendering realistic 3D models. These mathematical structures are used to
rotate, scale, and translate images efficiently.
In areas such as computer vision, an understanding of linear algebra enables the processing of image data as matrix transformations, essential for tasks like object recognition and 3D reconstruction.
In summary, the knowledge I’ve gained tightly links linear algebra to the practical aspects of computing that drive innovation. This connection underscores the value of a solid foundation in linear
algebra for any aspiring computer scientist. | {"url":"https://www.storyofmathematics.com/linear-algebra-in-computer-science/","timestamp":"2024-11-11T10:44:29Z","content_type":"text/html","content_length":"139242","record_id":"<urn:uuid:9a28d6a9-c84e-4584-a578-736df02a112c>","cc-path":"CC-MAIN-2024-46/segments/1730477028228.41/warc/CC-MAIN-20241111091854-20241111121854-00735.warc.gz"} |
Asia Pacific Wire & Cable Corporation Limited (APWC) DCF Excel Template
Asia Pacific Wire & Cable Corporation Limited (APWC) DCF Valuation
| | |
Real-Time Price ()
Market Cap A valuation method that multiplies the price of a company's shares by the total number of outstanding shares.
Revenue (ttm) The total amount of income generated by the sale of goods or services related to the company's primary operations.
Net Income (ttm) The company's earnings for a period net of operating costs, taxes and interest.
Shares Out Total number of common shares outstanding as of the latest date disclosed in a financial filing.
EPS (ttm) Company's net earnings or losses from continuing operations on a per diluted share basis.
PE Ratio The price-to-earnings (PE) ratio is the ratio between a company's stock price and earnings per share. It measures the price of a stock relative to its profits.
Dividend Yield Measures the cash returned to shareholders by a firm as a percentage of the price they pay for each share of stock.
Exchange Name of stock exchange where the trading item trades.
Avg Volume The average number of shares traded each day over the past 30 days.
Open The opening trade price over the trading day.
Previous Close The last closing price.
Beta A ratio that measures the risk or volatility of a company's share price in comparison to the market as a whole.
1 day delta The range between the high and low prices over the past day.
52 weeks The range between the high and low prices over the past 52 weeks.
Total Valuation
has a market cap or net worth of . The enterprise value is .
Market Cap (ttm) Market Capitalization
A valuation method that multiplies the price of a company's shares by the total number of outstanding shares.
Enterprise Value (ttm) Enterprise Value
Enterprise value measures the total value of a company's outstanding shares, adjusted for debt and levels of cash and short-term investments.
Enterprise Value = Market Cap + Total Debt - Cash & Equivalents - Short-Term Investments
Valuation Ratios
The trailing PE ratio is . 's PEG ratio is .
PE Ratio (ttm) PE Ratio
The price-to-earnings (P/E) ratio is a valuation metric that shows how expensive a stock is relative to earnings.
PE Ratio = Stock Price / Earnings Per Share
PS Ratio (ttm) PS Ratio
The price-to-sales (P/S) ratio is a commonly used valuation metric. It shows how expensive a stock is compared to revenue.
PS Ratio = Market Capitalization / Revenue
PB Ratio (ttm) PB Ratio
The price-to-book (P/B) ratio measures a stock's price relative to book value. Book value is also called Shareholders' equity.
PB Ratio = Market Capitalization / Shareholders' Equity
P/FCF Ratio (ttm) P/FCF Ratio
The price to free cash flow (P/FCF) ratio is similar to the P/E ratio, except it uses free cash flow instead of accounting earnings.
P/FCF Ratio = Market Capitalization / Free Cash Flow
PEG Ratio (ttm) PEG Ratio
The price/earnings to growth (PEG) ratio is calculated by dividing a company's PE ratio by its expected earnings growth.
PEG Ratio = PE Ratio / Expected Earnings Growth
Enterprise Valuation
The stock's EV/EBITDA ratio is , with a EV/FCF ratio of .
EV / Sales (ttm) EV / Sales Ratio
The enterprise value to sales (EV/Sales) ratio is similar to the price-to-sales ratio, but the price is adjusted for the company's debt and cash levels.
EV/Sales Ratio = Enterprise Value / Revenue
EV / EBITDA (ttm) EV / EBIT Ratio
The EV/EBITDA ratio measures a company's valuation relative to its EBITDA, or Earnings Before Interest, Taxes, Depreciation, and Amortization.
EV/EBITDA Ratio = Enterprise Value / EBITDA
EV / EBIT (ttm) EV/EBIT Ratio
The EV/EBIT is a valuation metric that measures a company's price relative to EBIT, or Earnings Before Interest and Taxes.
EV/EBIT Ratio = Enterprise Value / EBIT
EV / FCF (ttm) EV/FCF Ratio
The enterprise value to free cash flow (EV/FCF) ratio is similar to the price to free cash flow ratio, except the price is adjusted for the company's cash and debt.
EV/FCF Ratio = Enterprise Value / Free Cash Flow
Financial Efficiency
Return on equity (ROE) is and return on invested capital (ROIC) is .
Return on Equity (ROE) (ttm) Return on Equity (ROE)
Return on equity (ROE) is a profitability metric that shows how efficient a company is at using its equity (or "net" assets) to generate profits. It is calculated by dividing the company's net income
by the average shareholders' equity over the past 12 months.
ROE = (Net Income / Average Shareholders' Equity) * 100%
Return on Assets (ROA) (ttm) Return on Assets (ROA)
Return on assets (ROA) is a metric that measures how much profit a company is able to generate using its assets. It is calculated by dividing net income by the average total assets for the past 12
ROA = (Net Income / Average Total Assets) * 100%
Return on Capital (ROIC) (ttm) Return on Capital (ROIC)
Return on invested capital (ROIC) measures how effective a company is at investing its capital in order to increase profits. It is calculated by dividing the EBIT (Earnings Before Interest & Taxes)
by the average invested capital in the previous year.
ROIC = (EBIT / Average Invested Capital) * 100%
Asset Turnover Asset Turnover
The asset turnover ratio measures the amount of sales relative to a company's assets. It indicates how efficiently the company uses its assets to generate revenue.
Asset Turnover Ratio = Revenue / Average Assets
Inventory Turnover (ttm) Inventory Turnover
The inventory turnover ratio measures how many times inventory has been sold and replaced during a time period.
Inventory Turnover Ratio = Cost of Revenue / Average Inventory
Trailing 12 months gross margin is , with operating and profit margins of and .
Gross Margin (ttm) Gross Margin
Gross margin is the percentage of revenue left as gross profits, after subtracting cost of goods sold from the revenue.
Gross Margin = (Gross Profit / Revenue) * 100%
Operating Margin (ttm) Operating Margin
Operating margin is the percentage of revenue left as operating income, after subtracting cost of revenue and all operating expenses from the revenue.
Operating Margin = (Operating Income / Revenue) * 100%
Pretax Margin (ttm) Pretax Margin
Pretax margin is the percentage of revenue left as profits before subtracting taxes.
Pretax Margin = (Pretax Income / Revenue) * 100%
Profit Margin (ttm) Profit Margin
Profit margin is the percentage of revenue left as net income, or profits, after subtracting all costs and expenses from the revenue.
Profit Margin = (Net Income / Revenue) * 100%
EBITDA Margin (ttm) EBITDA Margin
EBITDA margin is the percentage of revenue left as EBITDA, after subtracting all expenses except interest, taxes, depreciation and amortization from revenue.
EBITDA Margin = (EBITDA / Revenue) * 100%
Income Statement
In the last 12 months, had revenue of and earned in profits. Earnings per share (EPS) was .
Revenue (ttm) Revenue
Revenue is the amount of money a company receives from its main business activities, such as sales of products or services. Revenue is also called sales.
Gross Profit (ttm) Gross Profit
Gross profit is a company’s profit after subtracting the costs directly linked to making and delivering its products and services.
Gross Profit = Revenue - Cost of Revenue
Operating Income (ttm) Operating Income
Operating income is the amount of profit in a company after paying for all the expenses related to its core operations.
Operating Income = Revenue - Cost of Revenue - Operating Expenses
Pretax Income (ttm) Pretax Income
Pretax income is a company's profits before accounting for income taxes.
Pretax Income = Net Income + Income Taxes
Net Income (ttm) Net Income
Net income is a company's accounting profits after subtracting all costs and expenses from the revenue. It is also called earnings, profits or "the bottom line"
Net Income = Revenue - All Expenses
EBITDA (ttm) EBITDA
EBITDA stands for "Earnings Before Interest, Taxes, Depreciation and Amortization." It is a commonly used measure of profitability.
EBITDA = Net Income + Interest + Taxes + Depreciation and Amortization
EBIT (ttm) EBIT
EBIT stands for "Earnings Before Interest and Taxes" and is a commonly used measure of earnings or profits. It is similar to operating income.
EBIT = Net Income + Interest + Taxes
Earnings Per Share (EPS) (ttm) EPS (Diluted)
Earnings per share is the portion of a company's profit that is allocated to each individual stock. Diluted EPS is calculated by dividing net income by "diluted" shares outstanding.
Diluted EPS = Net Income / Shares Outstanding (Diluted)
Financial Position
The company has a trailing 12 months (ttm) current ratio of , with a ttm Debt / Equity ratio of .
Current Ratio (ttm) Current Ratio
The current ratio is used to measure a company's short-term liquidity. A low number can indicate that a company will have trouble paying its upcoming liabilities.
Current Ratio = Current Assets / Current Liabilities
Quick Ratio (ttm) Quick Ratio
The quick ratio measure a company's short-term liquidity. A low number indicates that the company may have trouble paying its upcoming financial obligations.
Quick Ratio = (Cash + Short-Term Investments + Accounts Receivable) / Current Liabilities
Debt / Equity (ttm) Debt / Equity Ratio
The debt-to-equity ratio measures a company's debt levels relative to its shareholders' equity or book value. A high ratio implies that a company has a lot of debt.
Debt / Equity Ratio = Total Debt / Shareholders' Equity
Debt / EBIT (ttm) Debt / EBIT Ratio
The debt-to-EBIT ratio is a company's debt levels relative to its trailing twelve-month EBIT. A high ratio implies that debt is high relative to the company's earnings.
Debt / EBIT Ratio = Total Debt / EBIT (ttm)
Dividends & Yields
This stock pays an annual dividend of , which amounts to a dividend yield of .
Dividend Per Share (ttm) Dividend Per Share
Total amount paid to each outstanding share in dividends during the period.
Dividend Yield (ttm) Dividend Yield
The dividend yield is how much a stock pays in dividends each year, as a percentage of the stock price.
Dividend Yield = (Annual Dividends Per Share / Stock Price) * 100%
Earnings Yield (ttm) Earnings Yield
The earnings yield is a valuation metric that measures a company's profits relative to stock price, expressed as a percentage yield. It is the inverse of the P/E ratio.
Earnings Yield = (Earnings Per Share / Stock Price) * 100%
FCF Yield (ttm) FCF Yield
The free cash flow (FCF) yield measures a company's free cash flow relative to its price, shown as a percentage. It is the inverse of the P/FCF ratio.
FCF Yield = (Free Cash Flow / Market Cap) * 100%
Dividend Growth (YoY) Dividend Growth
The change in dividend payments per share, compared to the previous period.
Dividend Growth = ((Current Dividend / Previous Dividend) - 1) * 100%
Payout Ratio (ttm) Payout Ratio
The payout ratio is the percentage of a company's profits that are paid out as dividends. A high ratio implies that the dividend payments may not be sustainable.
Payout Ratio = (Dividends Per Share / Earnings Per Share) * 100%
Balance Sheet
The company has in cash and in debt, giving a net cash position of .
Cash & Cash Equivalents Cash & Cash Equivalents
Cash and cash equivalents is the sum of "Cash & Equivalents" and "Short-Term Investments." This is the amount of money that a company has quick access to, assuming that the cash equivalents and
short-term investments can be sold at a short notice.
Cash & Cash Equivalents = Cash & Equivalents + Short-Term Investments
Total Debt Total Debt
Total debt is the total amount of liabilities categorized as "debt" on the balance sheet. It includes both current and long-term (non-current) debt.
Total Debt = Current Debt + Long-Term Debt
Net Cash Net Cash / Debt
Net Cash / Debt is an indicator of the financial position of a company. It is calculated by taking the total amount of cash and cash equivalents and subtracting the total debt.
Net Cash / Debt = Total Cash - Total Debt
Book Value Shareholders' Equity
Shareholders’ equity is also called book value or net worth. It can be seen as the amount of money held by investors inside the company. It is calculated by subtracting all liabilities from all
Shareholders' Equity = Total Assets - Total Liabilities
Book Value Per Share (ttm) Book Value Per Share
Book value per share is the total amount of book value attributable to each individual stock. It is calculated by dividing book value (shareholders' equity) by the number of outstanding shares.
Book Value Per Share = Book Value / Shares Outstanding
Working Capital (ttm) Working Capital
Working capital is the amount of money available to a business to conduct its day-to-day operations. It is calculated by subtracting total current liabilities from total current assets.
Working Capital = Current Assets - Current Liabilities
Cash Flow
In the last 12 months, operating cash flow of the company was and capital expenditures , giving a free cash flow of .
Operating Cash Flow (ttm) Operating Cash Flow
Operating cash flow, also called cash flow from operating activities, measures the amount of cash that a company generates from normal business activities. It is the amount of cash left after all
cash income has been received, and all cash expenses have been paid.
Capital Expenditures (ttm) Capital Expenditures
Capital expenditures are also called payments for property, plants and equipment. It measures cash spent on long-term assets that will be used to run the business, such as manufacturing equipment,
real estate and others.
Free Cash Flow (ttm) Free Cash Flow
Free cash flow is the cash remaining after the company spends on everything required to maintain and grow the business. It is calculated by subtracting capital expenditures from operating cash flow.
Free Cash Flow = Operating Cash Flow - Capital Expenditures
FCF Per Share (ttm) Free Cash Flow Per Share
Free cash flow per share is the amount of free cash flow attributed to each outstanding stock.
FCF Per Share = Free Cash Flow / Shares Outstanding
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3.2.2 Expectation
If you have a collection of numbers $a_1,a_2,...,a_N$, their average is a single number that describes the whole collection. Now, consider a random variable $X$. We would like to define its average,
or as it is called in probability, its expected value or mean. The expected value is defined as the weighted average of the values in the range.
Expected value (= mean=average):
Let $X$ be a discrete random variable with range $R_X=\{x_1,x_2,x_3, ...\}$ (finite or countably infinite). The expected value of $X$, denoted by $EX$ is defined as $$EX=\sum_{x_k \in R_X} x_k P(X=
x_k)=\sum_{x_k \in R_X} x_k P_X(x_k).$$
To understand the concept behind $EX$, consider a discrete random variable with range $R_X=\{x_1,x_2,x_3, ...\}$. This random variable is a result of random experiment. Suppose that we repeat this
experiment a very large number of times $N$, and that the trials are independent. Let $N_1$ be the number of times we observe $x_1$, $N_2$ be the number of times we observe $x_2$, ...., $N_k$ be the
number of times we observe $x_k$, and so on. Since $P(X=x_k)=P_X(x_k)$, we expect that $$P_X(x_1)\approx \frac{N_1}{N},$$ $$P_X(x_2)\approx \frac{N_2}{N},$$ $$\hspace{10pt} . \hspace{20pt} . \hspace
{20pt} .$$ $$P_X(x_k)\approx \frac{N_k}{N},$$ $$\hspace{10pt} . \hspace{20pt} . \hspace{20pt} .$$ In other words, we have $N_k \approx N P_X(x_k)$. Now, if we take the average of the observed values
of $X$, we obtain
$\textrm{Average }$ $=\frac{N_1 x_1+N_2 x_2+N_3 x_3+...}{N}$
$\approx \frac{x_1 N P_X(x_1)+x_2N P_X(x_2)+x_3N P_X(x_3)+...}{N}$
$=x_1 P_X(x_1)+x_2 P_X(x_2)+x_3 P_X(x_3)+...$
Thus, the intuition behind $EX$ is that if you repeat the random experiment independently $N$ times and take the average of the observed data, the average gets closer and closer to $EX$ as $N$ gets
larger and larger. We sometimes denote $EX$ by $\mu_X$.
Different notations for expected value of $X$: $EX=E[X]=E(X)=\mu_X$.
Let's compute the expected values of some well-known distributions.
Let $X \sim Bernoulli(p)$. Find $EX$.
• Solution
□ For the Bernoulli distribution, the range of $X$ is $R_X=\{0,1\}$, and $P_X(1)=p$ and $P_X(0)=1-p$. Thus,
$EX$ $=0 \cdot P_X(0)+1 \cdot P_X(1)$
$=0 \cdot (1-p)+ 1 \cdot p$
For a Bernoulli random variable, finding the expectation $EX$ was easy. However, for some random variables, to find the expectation sum, you might need a little algebra. Let's look at another
Let $X \sim Geometric(p)$. Find $EX$.
• Solution
□ For the geometric distribution, the range is $R_X=\{1,2,3,... \}$ and the PMF is given by $$P_X(k) = q^{k-1}p, \hspace{20pt} \text{ for } k=1,2,...$$ where, $0 < p < 1$ and $q=1-p$. Thus, we
can write
$EX$ $=\sum_{x_k \in R_X} x_k P_X(x_k)$
$=\sum_{k=1}^{\infty} k q^{k-1}p$
$=p\sum_{k=1}^{\infty} k q^{k-1}$.
Now, we already know the geometric sum formula $$\sum_{k=0}^{\infty} x^k= \frac{1}{1-x}, \hspace{20pt} \textrm{ for } |x| < 1.$$ But we need to find a sum $\sum_{k=1}^{\infty} k q^{k-1}$.
Luckily, we can convert the geometric sum to the form we want by taking derivative with respect to $x$, i.e., $$\frac{d}{dx} \sum_{k=0}^{\infty} x^k= \frac{d}{dx} \frac{1}{1-x}, \hspace{20pt}
\textrm{ for } |x| < 1.$$ Thus, we have $$\sum_{k=0}^{\infty} k x^{k-1}= \frac{1}{(1-x)^2}, \hspace{20pt} \textrm{ for } |x| < 1.$$ To finish finding the expectation, we can write
$EX$ $=p\sum_{k=1}^{\infty} k q^{k-1}$
$=p \frac{1}{(1-q)^2}$
$=p \frac{1}{p^2}$
So, for $X \sim Geometric(p)$, $EX=\frac{1}{p}$. Note that this makes sense intuitively. The random experiment behind the geometric distribution was that we tossed a coin until we observed
the first heads, where $P(H)=p$. Here, we found out that on average you need to toss the coin $\frac{1}{p}$ times in this experiment. In particular, if $p$ is small (heads are unlikely), then
$\frac{1}{p}$ is large, so you need to toss the coin a large number of times before you observe a heads. Conversely, for large $p$ a few coin tosses usually suffices.
Let $X \sim Poisson(\lambda)$. Find $EX$.
• Solution
□ Before doing the math, we suggest that you try to guess what the expected value would be. It might be a good idea to think about the examples where the Poisson distribution is used. For the
Poisson distribution, the range is $R_X=\{0,1,2,\cdots \}$ and the PMF is given by $$P_X(k) = \frac{e^{-\lambda} \lambda^k}{k!}, \hspace{20pt} \text{ for } k=0,1,2,...$$ Thus, we can write
$EX$ $=\sum_{x_k \in R_X} x_k P_X(x_k)$
$= \sum_{k=0}^{\infty} k \frac{e^{-\lambda} \lambda^k}{k!}$
$=e^{-\lambda} \sum_{k=1}^{\infty} \frac{ \lambda^k}{(k-1)!}$
$=e^{-\lambda} \sum_{j=0}^{\infty} \frac{\lambda^{(j+1)}}{j!}$ $(\textrm{ by letting }j=k-1)$
$=\lambda e^{-\lambda} \sum_{j=0}^{\infty} \frac{ \lambda^j}{j!}$
$=\lambda e^{-\lambda} e^{\lambda}$ $(\textrm{ Taylor series for } e^{\lambda})$
So the expected value is $\lambda$. Remember, when we first talked about the Poisson distribution, we introduced its parameter $\lambda$ as the average number of events. So it is not
surprising that the expected value is $EX=\lambda$.
Before looking at more examples, we would like to talk about an important property of expectation, which is linearity. Note that if $X$ is a random variable, any function of $X$ is also a random
variable, so we can talk about its expected value. For example, if $Y=aX+b$, we can talk about $EY=E[aX+b]$. Or if you define $Y=X_1+X_2+\cdots+X_n$, where $X_i$'s are random variables, we can talk
about $EY=E[X_1+X_2+\cdots+X_n]$. The following theorem states that expectation is linear, which makes it easier to calculate the expected value of linear functions of random variables.
Expectation is linear:
We have
• $E[aX+b]=aEX+b$, for all $a,b \in \mathbb{R}$;
• $E[X_1+X_2+\cdots+X_n]=EX_1+EX_2+\cdots+EX_n$, for any set of random variables $X_1, X_2,\cdots,X_n$.
We will prove this theorem later on in Chapter 5, but here we would like to emphasize its importance with an example.
Let $X \sim Binomial(n,p)$. Find $EX$.
• Solution
□ We provide two ways to solve this problem. One way is as before: we do the math and calculate $EX=\sum_{x_k \in R_X} x_k P_X(x_k)$ which will be a little tedious. A much faster way would be
to use linearity of expectation. In particular, remember that if $X_1, X_2, ...,X_n$ are independent $Bernoulli(p)$ random variables, then the random variable $X$ defined by $X=
X_1+X_2+...+X_n$ has a $Binomial(n,p)$ distribution. Thus, we can write
$EX$ $=E[X_1+X_2+\cdots+X_n]$
$=EX_1+EX_2+\cdots+EX_n$ $\hspace{20pt} \textrm{by linearity of expectation}$
We will provide the direct calculation of $EX=\sum_{x_k \in R_X} x_k P_X(x_k)$ in the Solved Problems section and as you will see it needs a lot more algebra than above. The bottom line is
that linearity of expectation can sometimes make our calculations much easier. Let's look at another example.
Let $X \sim Pascal(m,p)$. Find $EX$. (Hint: Try to write $X=X_1+X_2+\cdots+X_m$, such that you already know $EX_i$.)
• Solution
□ We claim that if the $X_i$'s are independent and $X_i \sim Geometric(p)$, for $i=1$, $2$, $\cdots$, $m$, then the random variable $X$ defined by $X=X_1+X_2+\cdots+X_m$ has $Pascal(m,p)$. To
see this, you can look at Problem 5 in Section 3.1.6 and the discussion there. Now, since we already know $EX_i=\frac{1}{p}$, we conclude
$EX$ $=E[X_1+X_2+\cdots+X_m]$
$=EX_1+EX_2+\cdots+EX_m$ $\hspace{20pt} \textrm{by linearity of expectation}$
Again, you can try to find $EX$ directly and as you will see, you need much more algebra compared to using the linearity of expectation.
The print version of the book is available on Amazon.
Practical uncertainty: Useful Ideas in Decision-Making, Risk, Randomness, & AI | {"url":"https://www.probabilitycourse.com/chapter3/3_2_2_expectation.php","timestamp":"2024-11-09T12:38:43Z","content_type":"text/html","content_length":"23827","record_id":"<urn:uuid:fb90d1ba-c946-4b6f-b765-61c8629a4905>","cc-path":"CC-MAIN-2024-46/segments/1730477028118.93/warc/CC-MAIN-20241109120425-20241109150425-00746.warc.gz"} |
NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.5
Here you will find Chapter 3 Playing with Numbers Exercise 3.5 Class 6 Maths NCERT Solutions which will help you in understanding the basics of the chapter. The answers provided here are detailed so
you can understand every concepts given in the question easily and also practice yourself and check later. NCERT Solutions are updated as per the latest marking scheme released by CBSE. It will
prepare you for higher classes and also improve your marks in the examinations. | {"url":"https://www.studyrankers.com/2021/02/ncert-solutions-for-class106-maths-playing-with-numbers-exercise-3.5.html","timestamp":"2024-11-06T14:06:49Z","content_type":"application/xhtml+xml","content_length":"293644","record_id":"<urn:uuid:93af06a0-84e0-4b24-b625-699785c892ba>","cc-path":"CC-MAIN-2024-46/segments/1730477027932.70/warc/CC-MAIN-20241106132104-20241106162104-00468.warc.gz"} |
Martini, Grasiela January 2013 (has links)
Neste trabalho discutimos a semissimplicidade de álgebras de Hopf finito-dimensionais e construímos o Duplo de Drinfeld D(H) de uma tal álgebra H. Além disso, apresentamos um resultado mostrando a
equivalência entre as categorias de representações dos módulos sobre D(H) e dos módulos de Yetter-Drinfeld sobre Hcop. Como consequência deste estudo, apresentamos um resultado que caracteriza uma
41 álgebra de Hopf quase triangular. / In this work we discuss the semisimplicity of some finite-dimensional Hopf Algebras and we set up the Drinfel’d double D(H) of such an algebra H. In addiction,
we present a result showing the equivalence between the representation category of modules over D(H) and the Yetter-Drinfeld modules over Hcop. As a consequence of this, we present a result that
features a quasitriangular Hopf algebra.
Algebras de hopf
Cardoso, Kauê da Rosa January 2013 (has links)
O objetivo deste trabalho é mostrar que o conjunto de todas G-super palavras monótonas restritas escritas com super letras duras forma uma base para a álgebra de Hopf H, onde H é gerada por um
42 conjunto skew-primitivo semi-invariante {a1, ..., an} e um grupo abeliano G de todos elementos grouplike. / The objective of this work is to show that the set of all monotonic restricted
G-super-words written with hard super-letters form a basis for the Hopf algebra H, where H is generated by a skew-primitive semi-invariant set {a1, ..., an} and an abelian group G of all
group-like elements.
Algebras de hopf
Behrendt, Darren Robin 24 January 2012 (has links)
43 D.Phil.
Banach algebras
Renison, Martin 27 June 2008 (has links)
44 Dr. R.M Brits
Banach algebras
Cheung, Wai-Shun 07 May 2018 (has links)
In this dissertation, we study certain types of linear mappings on triangular algebras. Triangular algebras are algebras whose elements can be written in the form of 2 x 2 matrices [special
characters omitted]where a ∈ A, b ∈ B, m ∈ M and where A, B are algebras and M is a bimodule. Many widely studied algebras, such as upper triangular matrix algebras and nest algebras, can be
viewed as triangular algebras. This dissertation is divided into five chapters. The first chapter is a general account of the basics of triangular algebras, including the unitization of nonunital
triangular algebras and the structure of the centre of triangular algebras, as well as a brief introduction to some well-known examples of triangular algebras. In Chapter 2, we study the general
structure of derivations on triangular algebras and obtain some results on the first cohomology groups of triangular algebras. The first cohomology group of an algebra is the quotient space of the
45 space of all derivations over the space of all inner derivations, and it is always a main tool in the research of derivations. In addition, we consider the problem of automatic continuity of
derivations in the last section of this chapter. In Chapter 3, we consider sufficient conditions on a triangular algebra so that every Lie derivation is a sum of a derivation and a linear map
whose image lies in the centre of the triangular algebra. In Chapter 4, we consider sufficient conditions for every commuting map on a triangular algebra to be a sum of a map of the form x ↦ ax
and a map whose image lies in the centre of the triangular algebra. In the final chapter, we are concerned with the automorphisms of triangular algebras. The study of automorphism is a most
important way to understand the underlying structure of an algebra. We deduce some results on the Skolem-Noether groups, or the outer automorphism groups, of triangular algebras and apply those
results to generalize some known results about automorphisms on a triangular matrix algebras. / Graduate
Algebras, Linear
Kone, Namadzavho Bernard 30 March 2009 (has links)
46 M.Sc.
Banach algebras
Weitz, Craig Stewart 25 May 2010 (has links)
47 M.Sc.
Banach algebras
Edwards, C. M. January 1966 (has links)
48 No description available.
Group algebras
Vaughan-Lee, Michael January 1968 (has links)
49 No description available.
Lie algebras
Chao, Frances Yen-Yen January 1973 (has links)
In this thesis we show that it is impossible to define a Hopf multiplication on the wedge X V Y of a connected finite CW-complex X with an arbitrary connected CW-complex Y, provided neither X nor
50 Y homotopy equivalent to a point. We also show by an example, that the above is no longer true, if one does not require that X is a finite CW-complex. / Science, Faculty of / Mathematics,
Department of / Graduate
Hopf algebras. | {"url":"http://search.ndltd.org/search.php?q=subject%3A%22Algebras%22&start=40","timestamp":"2024-11-06T20:18:23Z","content_type":"text/html","content_length":"67498","record_id":"<urn:uuid:11a204e2-0d27-426e-9d8c-382f7cb77c67>","cc-path":"CC-MAIN-2024-46/segments/1730477027942.47/warc/CC-MAIN-20241106194801-20241106224801-00512.warc.gz"} |
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AMS Chelsea Publishing: An Imprint of the American Mathematical Society
Hardcover ISBN: 978-0-8218-3436-7
Product Code: CHEL/346.H
List Price: $69.00
MAA Member Price: $62.10
AMS Member Price: $62.10
eBook ISBN: 978-1-4704-2997-3
Product Code: CHEL/346.H.E
List Price: $65.00
MAA Member Price: $58.50
AMS Member Price: $58.50
Hardcover ISBN: 978-0-8218-3436-7
eBook: ISBN: 978-1-4704-2997-3
Product Code: CHEL/346.H.B
List Price: $134.00 $101.50
MAA Member Price: $120.60 $91.35
AMS Member Price: $120.60 $91.35
Click above image for expanded view
Knots and Links
AMS Chelsea Publishing: An Imprint of the American Mathematical Society
Hardcover ISBN: 978-0-8218-3436-7
Product Code: CHEL/346.H
List Price: $69.00
MAA Member Price: $62.10
AMS Member Price: $62.10
eBook ISBN: 978-1-4704-2997-3
Product Code: CHEL/346.H.E
List Price: $65.00
MAA Member Price: $58.50
AMS Member Price: $58.50
Hardcover ISBN: 978-0-8218-3436-7
eBook ISBN: 978-1-4704-2997-3
Product Code: CHEL/346.H.B
List Price: $134.00 $101.50
MAA Member Price: $120.60 $91.35
AMS Member Price: $120.60 $91.35
• AMS Chelsea Publishing
Volume: 346; 1976; 439 pp
MSC: Primary 57
Rolfsen's beautiful book on knots and links can be read by anyone, from beginner to expert, who wants to learn about knot theory. Beginners find an inviting introduction to the elements of
topology, emphasizing the tools needed for understanding knots, the fundamental group and van Kampen's theorem, for example, which are then applied to concrete problems, such as computing knot
groups. For experts, Rolfsen explains advanced topics, such as the connections between knot theory and surgery and how they are useful to understanding three-manifolds.
Besides providing a guide to understanding knot theory, the book offers “practical” training. After reading it, you will be able to do many things: compute presentations of knot groups, Alexander
polynomials, and other invariants; perform surgery on three-manifolds; and visualize knots and their complements. It is characterized by its hands-on approach and emphasis on a visual, geometric
Rolfsen offers invaluable insight and strikes a perfect balance between giving technical details and offering informal explanations. The illustrations are superb, and a wealth of examples are
Now back in print by the AMS, the book is still a standard reference in knot theory. It is written in a remarkable style that makes it useful for both beginners and researchers. Particularly
noteworthy is the table of knots and links at the end. This volume is an excellent introduction to the topic and is suitable as a textbook for a course in knot theory or 3-manifolds.
Other key books of interest on this topic available from the AMS are The Shoelace Book: A Mathematical Guide to the Best (and Worst) Ways to Lace your Shoes and The Knot Book.
Advanced undergraduates, graduate students, and research mathematicians interested in knot theory and its applications to low-dimensional topology.
□ Chapters
□ Chapter 1. Introduction
□ Chapter 2. Codimension one and other matters
□ Chapter 3. The fundamental group
□ Chapter 4. Three-dimensional PL geometry
□ Chapter 5. Seifert surfaces
□ Chapter 6. Finite cyclic coverings and the torsion invariants
□ Chapter 7. Infinite cyclic coverings and the Alexander invariant
□ Chapter 8. Matrix invariants
□ Chapter 9. 3-manifolds and surgery on links
□ Chapter 10. Foliations, branched covers, fibrations and so on
□ Chapter 11. A higher-dimensional sampler
□ Appendix A. Covering spaces and some algebra in a nutshell
□ Appendix B. Dehn’s lemma and the loop theorem
□ Appendix C. Table of knots and links
□ ...a gem and a classic. Every mathematics library should own a copy and every mathematician should read at least some of it. The writing is clear and engaging, while the choice of examples is
genius...Rolfsen's book continues to be a beautiful introduction to some beautiful ideas.
Scott A. Taylor, MAA Reviews
• Permission – for use of book, eBook, or Journal content
• Book Details
• Table of Contents
• Reviews
• Requests
Volume: 346; 1976; 439 pp
MSC: Primary 57
Rolfsen's beautiful book on knots and links can be read by anyone, from beginner to expert, who wants to learn about knot theory. Beginners find an inviting introduction to the elements of topology,
emphasizing the tools needed for understanding knots, the fundamental group and van Kampen's theorem, for example, which are then applied to concrete problems, such as computing knot groups. For
experts, Rolfsen explains advanced topics, such as the connections between knot theory and surgery and how they are useful to understanding three-manifolds.
Besides providing a guide to understanding knot theory, the book offers “practical” training. After reading it, you will be able to do many things: compute presentations of knot groups, Alexander
polynomials, and other invariants; perform surgery on three-manifolds; and visualize knots and their complements. It is characterized by its hands-on approach and emphasis on a visual, geometric
Rolfsen offers invaluable insight and strikes a perfect balance between giving technical details and offering informal explanations. The illustrations are superb, and a wealth of examples are
Now back in print by the AMS, the book is still a standard reference in knot theory. It is written in a remarkable style that makes it useful for both beginners and researchers. Particularly
noteworthy is the table of knots and links at the end. This volume is an excellent introduction to the topic and is suitable as a textbook for a course in knot theory or 3-manifolds.
Other key books of interest on this topic available from the AMS are The Shoelace Book: A Mathematical Guide to the Best (and Worst) Ways to Lace your Shoes and The Knot Book.
Advanced undergraduates, graduate students, and research mathematicians interested in knot theory and its applications to low-dimensional topology.
• Chapters
• Chapter 1. Introduction
• Chapter 2. Codimension one and other matters
• Chapter 3. The fundamental group
• Chapter 4. Three-dimensional PL geometry
• Chapter 5. Seifert surfaces
• Chapter 6. Finite cyclic coverings and the torsion invariants
• Chapter 7. Infinite cyclic coverings and the Alexander invariant
• Chapter 8. Matrix invariants
• Chapter 9. 3-manifolds and surgery on links
• Chapter 10. Foliations, branched covers, fibrations and so on
• Chapter 11. A higher-dimensional sampler
• Appendix A. Covering spaces and some algebra in a nutshell
• Appendix B. Dehn’s lemma and the loop theorem
• Appendix C. Table of knots and links
• ...a gem and a classic. Every mathematics library should own a copy and every mathematician should read at least some of it. The writing is clear and engaging, while the choice of examples is
genius...Rolfsen's book continues to be a beautiful introduction to some beautiful ideas.
Scott A. Taylor, MAA Reviews
Permission – for use of book, eBook, or Journal content
You may be interested in...
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Algebras and Hilbert spaces from gravitational path integrals (Nov. 30, 2023)
Date: Nov. 30 (Thur.), 2023
In this talk, I will describe a construction of Hilbert spaces and von Neumann algebras from any UV-completion of asymptotically anti-de Sitter quantum gravity with a Euclidean path integral
satisfying a simple and familiar set of axioms. We consider a quantum context in which a standard Lorentz-signature classical bulk limit wouldhave Cauchy slices with two asymptotic boundaries (left
and right), both of which are compact manifolds without boundary. Our main result is then that the quantum gravity path integral defines (left and right) type I von Neumann algebras of observables
acting respectively at the left and right boundaries, such that the two algebras are commutants. The path integral also defines entropies on the von Neumann algebras. The entropies can also be
written in terms of standard density matrices and standard Hilbert space traces. Furthermore, in appropriate semiclassical limits our entropies are computed by the RT-formula with quantumcorrections.
Our work thus provides a Hilbert space interpretation of the Ryu-Takayanagi entropy. Since our axioms do not restrict UV bulk structures, they may be expected to hold equally well for successful
formulations of string field theory, spin-foam models, or any other approach to constructing a UV-complete theory of gravity. | {"url":"https://kits.ucas.ac.cn/index.php/events/seminars/586-algebras-and-hilbert-spaces-from-gravitational-path-integrals","timestamp":"2024-11-04T18:47:11Z","content_type":"text/html","content_length":"18124","record_id":"<urn:uuid:c1077de5-ff77-4b1e-bdcd-a323ba05cf25>","cc-path":"CC-MAIN-2024-46/segments/1730477027838.15/warc/CC-MAIN-20241104163253-20241104193253-00890.warc.gz"} |
Factors and Multiples: My First Unit in 6th Grade Math
As I begin to prepare for back to school, I am reminded that teaching factors and multiples at the beginning of the year aids in teaching future units.
Check Out How I Have Organized My First Unit in 6th Grade Math.
Chapter 1: Factors and Multiples
In my first chapter, I start by talk to my students about factors and factor pairs. I show them how to find all the factors of a number by finding factor pairs using a factor rainbow. This leads to a
conversation about prime. composite, and square numbers. By looking at all the factors of a number, students can identify if that number is prime, composite, square, or a combination of these.
Finally, in chapter 1, I introduce multiples.
Chapter 2: Common Factors and Common Multiples
In chapter two, I take the students knowledge from factors and multiples, and start discussing common factors and common multiples. I start with common factors and discuss that the number 1 is ALWAYS
a common factor of any two numbers. I also have the students start thinking about how they will use these concepts in the real world. I show them that common factors are used to solve "sharing
problems". I then move on to common multiples and we discuss how to find common multiples. I also show them that common multiples are used to solve "cycle" problems or problems that involve
repetition. Part of this chapter involves word problems for my students to practice these skills.
Mid-Unit Quiz
Here is where I check my students progress so far in this unit. I give my students a quiz review to complete and then I give them a quiz that I use to assess my students.
Chapter 3: Prime Factorization, Distributive Property, and Order of Operations
Chapter 3 is where I start to introduce some new topics. First, I introduce prime factorization. I show my students that by using factor pairs, they can reduce a number down to a factor string of
only prime numbers. This is where I introduce short hand notation and exponents. Here is a link to my
FREE prime factorization lesson
I then move on to the distributive property. I introduce this by having students discover that there are two ways to find the area of a big rectangle when it is created by two smaller rectangles.
I end this unit by teaching order of operations which includes parentheses and exponents. I also give my students a couple of pages of word problems to help them discover which operation they should
use when solving story problems.
Unit Test
Here is where I take a final assessment over my entire unit.
In Conclusion
This unit has three chapters. Each chapter has between 3-4 lesson in it. My lessons include notes and practice worksheet that concentrate on a particular topic or standard. This unit has a quiz
review and a quiz that covers the first two chapters and a test that covers the entire unit. Answer keys are included! If you want to check out my first unit click on
. I have created it in two different versions to suit your needs.
I have also created an ENTIRE curriculum for 6th grade math. I have made this in two versions as well. If you are interested in looking at the entire 6th grade curriculum click on the
GOOGLE version
or the
PDF version
For the Google versions, I assign the lessons with my students using Google Classroom. I have put text boxes on the Google Slides to make it easier for my students to type their answers. I have also
made them 8.5x11 so they can easily be printed off as well. Here are
3 reasons why I love using Google in my math classroom | {"url":"https://www.mitchellsmathematicians.com/2019/07/factors-and-multiples-my-first-unit-in.html","timestamp":"2024-11-08T01:28:12Z","content_type":"application/xhtml+xml","content_length":"186597","record_id":"<urn:uuid:ae0e3f05-ba18-4cc0-8d2d-b9bb41e29511>","cc-path":"CC-MAIN-2024-46/segments/1730477028019.71/warc/CC-MAIN-20241108003811-20241108033811-00055.warc.gz"} |
UPSEE-B-TECH | 2022 Free Mock Test Papers,
Online Test Series,
Online Preparation,
Practice Set,
Applications Forms,
Kindly Note : TestBag now has exclusive Microsite for UPSEE 2020 Engineering Entrance Test for Admission to First Year of BTech. Please click on upsee.testbag.com for one stop on everything about
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The UPSEE 2020 Engineering Entrance Test Question paper consists of objective multiple type questions.
The UPSEE 2020 Engineering Entrance Test Question paper consists of Physics, Chemistry and Mathematics 50 objective type questions, each of Physics, Chemistry and Mathematics with a total of 150
questions and total of 600 Marks
Section A PHYSICS
Measurement : Dimensional analysis and error estimation, dimensional compatibility and significant figures.
Motion in one dimension : Average velocity, instantaneous velocity, one-dimensional motion with constant accelerations, freely falling bodies.
Laws of Motion : Force and inertia, Newtons laws of motion, and their significance.
Motion in two dimensions : Projectile motion, uniform circular motion, tangential and radial acceleration in curve-linear motion, relative motion and relative acceleration.
Work, Power and Energy : Work done by a constant and variable forces, kinetic and potential energy, power, Conservative and non-conservative forces, conservation of energy, gravitational energy, work
energy theorem, potential energy stored in a spring.
Linear Momentum & collisions : Linear momentum & impulse, conservation of linear momentum for two particle system, collisions, collision in one dimension, collision in two dimension, rocket
Rotation of a rigid body about a fixed axis : Angular velocity and angular acceleration, rotational kinematics, rotational motion with constant angular acceleration relationship between angular and
linear quantities, rotational energy, moment of inertia for a ring, rod, spherical shell, sphere and plane lamina, torque and angular acceleration, work and energy in rotational motion, rolling
motion of a solid sphere and cylinder.
Gravitation : Gravitational field, Keplers laws and motion of planets, planetary and satellite motion, geostationary satellite.
Oscillatory motion : Harmonic motion, oscillatory motion of mass attached to a spring, kinetic & potential energy, Time Period of a simple pendulum, comparing simple and harmonic motion with uniform
circular motion, forced oscillations, damped oscillations and resonance.
Click Here For UPSEE Online Free Mock test Series
Mechanics of solids and fluids : States of matter youngs modulus, bulk modulus, shear modulus of rigidity, variations of pressure with depth, Buoyant forces and Archimedes principle, Pascals law,
Bernoullis theorem and its application, surface energy, surface tension, angle of contact, capillary rise, coefficient of viscosity, viscous force, terminal velocity, Stokes law, stream line motion,
Reynolds numbers.
Heat and thermodynamics : First law of thermodynamics, specific heat of an ideal gas at constant volume and constant pressure, relation between them, thermodynamics process (reversible, irreversible,
isothermal, adiabatic), second law of thermodynamics, concept of entropy and concept of absolute scale, efficiency of a Carnot engine, thermal conductivity, Newtons law of cooling, black body
radiation, Wiens displacement law, Stefans law.
Wave : Wave motion, phase, amplitude and velocity of wave, Newtons formula for longitudinal waves, propagation of sound waves in air, effect of temperature and pressure on velocity of sound, Laplaces
correction, Principle of superposition, formation of standing waves, standing waves in strings and pipes, beats, Dopplers effect.
Electrostatics : Coulombs law, electric field and potential due to point charge, dipole and its field along the axis and perpendicular to axis, electric flux, Gausss theorem and its applications to
find the field due to infinite sheet of charge, and inside the hallow conducting sphere, capacitance, parallel plate capacitor with air and dielectric medium between the Plates, series and parallel
combination of capacitors, energy of a capacitor, displacement currents.
Current Electricity : Concept of free and bound electrons, drift velocity and mobility, electric current, Ohms law, resistivity, conductivity, temperature dependency of resistance, resistance in
series and parallel combination, Kirchhoffs law and their application to network of resistances, principle of potentiometer, effect of temperature on resistance and its application.
Magnetic Effect of Current : Magnetic field due to current, Biot-Savarts law, magnetic field due to solenoid, motion of charge in a magnetic field, force on a current carrying conductors and torque
on current loop in a magnetic field, magnetic flux, forces between two parallel current carrying conductors, moving coil galvanometer and its conversion into ammeter and voltmeter.
Magnetism in Matter : The magnetization of substance due to orbital and spin motions of electrons, magnetic moment of atoms, diamagnetism, paramagnetism, ferromagnetism, earths magnetic field and its
components and their measurement.
Electro magnetic induction : Induced e.m.f., Faradays laws,Lenzs law,electromagnetic induction, self and mutual induction, B-H curve, hysteresis loss and its importance, eddycurrents.
Ray Optics and optical instruments : Sources of light, luminous intensity, luminous flux, illuminance, photometry, wave nature of light, Huygens theory for propagation of light and rectilinear
propagation of light, reflection of light , total internal reflection, reflection and refraction at spherical surfaces, focal length of a combination of lenses, spherical and chromatic aberration and
their removal, refraction and dispersion of light due to a prism, simple and compound microscope, reflecting and refracting telescope, magnifying power and resolving power.
Wave Optics : Coherent and incoherent sources of light, interference, youngs double slit experiment diffraction due to a single slit, linearly polarized light, Polaroid.
Modern Physics : Photo-electric equation, matterwaves, quantization, Plancks hypothesis, Bohrs model of hydrogen atom and its spectra, ionization potential, Rydberg constant, solar spectrum and
Fraunhofer lines, fluorescence and phosphorescence, X-Rays and their productions, characteristic and continuous spectra.
Nuclear Instability, radioactive decay laws, Emission of Alpha, Beta and Gamma Rays, Mass - defect, Mass Energy equivalence, Nuclear Fission, Nuclear Reactors, Nuclear Fusion. Classification of
conductors, Insulators and semiconductors on the basis of energy bands in solids, PN junction, PN Diode, junction Transistors, Transistor as an amplifier and Oscillator.
Principles of Logic Gates ( AND, OR and NOT ) Analog Vs Digital communication, Difference between Radio and television, Signal propagation, Principle of LASER and MASER, Population Inversion,
Spontaneous and stimulated Emission
Section B CHEMISTRY
Atomic Structure : Bohrs concept. Quantum numbers, Electronic configuration, molecular orbital theory for homo-nuclear molecules, Paulis exclusion principle.
Chemical Bonding : Electrovalency, co-valency, hybridization involving s, p and d orbitals hydrogen bonding.
Redox Reactions : Oxidation number, oxidising and reducing agents, balancing of equations.
Chemical Equilibrium and Kinetics : Equilibrium constant (for gaseous system only) Le Chateliers principle, ionic equilibrium, Ostwalds dilution law, hydrolysis, pH and buffer solution, solubility
product, common-ion effect, rate constant and first order reaction
Acid-Base Concepts : Bronsted Lowry & Lewis.
Electrochemistry : Electrode potential and electro-chemical series.
Catalysis : Types and applications.
Colloids : Types and preparation, Brownian movement, Tyndall effect, coagulation and peptization.
Colligative Properties of Solution : Lowering of vapor pressure, Osmotic pressure, depression of freezing point, elevation of boiling point, determination of molecular weight.
Periodic Table : Classification of elements on the basis of electronic configuration, properties of s,p and d block elements, ionization potential, electronegativity & electron affinity.
Preparation and Properties of the following : Hydrogen peroxide. copper sulfate, silver nitrate, plaster of paris, borax, Mohrs salt, alums, white and red lead, microcosmic salt and bleaching powder,
sodium thiosulfate.
Thermo-chemistry : Exothermic & endothermic reactions Heat of reaction, Heat of combustion & formation, neutralization, Hesss law.
General Organic Chemistry : Shape of organic compounds, Inductive effect, mesomeric effect, electrophiles & nucleophiles, Reaction intermediates: carbonium ion, carbanions & free radical, Types of
organic reactions, Cannizzaro Friedel Craft, Perkin, Aldol condensation.
Isomerism : Structural, Geometrical & Optical IUPAC : Nomenclature of simple organic compounds.
Polymers : Addition & condensation polymers Carbohydrates: Monosaccharides.
Preparation and Properties Of the Followings : Hydrocarbons, monohydric alcohols, aldehydes, ketones, monocarboxylic acids, primary amines, benzene, nitrobenzene, aniline, phenol, benzaldehyde,
benzoic acid, Grignard Reagent.
Solid State : Structure of simple ionic compounds, Crystal imperfections (point defects only), Born-Haber cycle
Petroleum : Important industrial fractions, cracking, octane number, anti-knocking compounds.
Section C MATHEMATICS
Algebra : Sets relations & functions, De-Morgans Law, Mapping Inverse relations, Equivalence relations, Peanos axioms, Definition of rationals and integers through equivalence relation, Indices and
surds, Solutions of simultaneous and quadratic equations, A.P., G.P. and H.P., Special sums i.e. ∑n2 and ∑n3(n∑N ), Partial fraction, Binomial theorem for any index, exponential series, Logarithm and
Logarithmic series. Determinants and their use in solving simultaneous linear equations, Matrices, Algebra of matrices, Inverse of a matrix, Use of matrix for solving equations.
Probability : Definition, Dependent and independent events, Numerical problem on addition and multiplication, theorem of probability.
Trigonometry : Identities, Trigonometric equations, properties of triangles, solution of triangles, heights and distances, Inverse function, Complex numbers and their properties, Cube roots of unity,
De-Moivres theorem.
Co-ordinate Geometry : Pair of straight lines, Circles, General equation of second degree, parabola, ellipse and hyperbola, tracing of conics.
Calculus : Limits & continuity of functions, Differentiation of function of function, tangents & normal, Simple examples of Maxima & Minima, Indeterminate forms, Integration of function by parts, by
substitution and by partial fraction, definite integral, application to volumes and surfaces of frustums of sphere, cone and cylinder. Differential equations of first order and of first degree.
Vectors : Algebra of vectors, scalar and vector products of two and three vectors and their applications.
Dynamics : Velocity, composition of velocity, relative velocity, acceleration, composition of accelerations, Motion under gravity, Projectiles, Laws of motion, Principles of conservation of momentum
and energy, direct impact of smooth bodies.
Statics : Composition of coplanar, concurrent and parallel forces moments and couples resultant of set of coplanar forces and condition of equilibrium, determination of centroid in simple cases,
Problems involving friction.
Kindly Note : The information provided here is just indicative information and is provided on "as is" and "as available" basis . We make no claims on accuracy and reliability of the information. For
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Grid Method Multiplication Worksheet
Math, especially multiplication, forms the keystone of numerous scholastic self-controls and real-world applications. Yet, for several students, understanding multiplication can posture a difficulty.
To resolve this hurdle, instructors and parents have actually embraced an effective tool: Grid Method Multiplication Worksheet.
Introduction to Grid Method Multiplication Worksheet
Grid Method Multiplication Worksheet
Grid Method Multiplication Worksheet -
Using the grid method to solve multiplication problems This worksheet has a focus on the 8 times tables
Students can use math worksheets to master a math skill through practice in a study group or for peer tutoring Use the buttons below to print open or download the PDF version of the 2 digit by 2
digit Multiplication with Grid Support Including Regrouping A math worksheet The size of the PDF file is 181300 bytes
Importance of Multiplication Practice Understanding multiplication is critical, laying a strong structure for innovative mathematical principles. Grid Method Multiplication Worksheet use structured
and targeted method, fostering a deeper comprehension of this basic math procedure.
Evolution of Grid Method Multiplication Worksheet
Grid Method Multiplication TMK Education
Grid Method Multiplication TMK Education
The grid method of multiplication helps children separate multiplication questions into smaller chunks allowing them to then simply add their individual answers together to find the correct answer
The questions on these multiplication grid worksheets cover multiplying 2 digit numbers by 1 and 2 digit numbers
Once you have found the missing numbers add all the numbers in the grid using the column method to calculate the answer 1 2 3 4 Activity 1 Work out the following multiplications using the grid method
1 16 x 3 5 19 x 8 9 125 x 3 13 828 x 6 2 17 x 5 6 73 x 3 10 253 x 6 14 901 x 9 3 23 x 4
From conventional pen-and-paper workouts to digitized interactive layouts, Grid Method Multiplication Worksheet have evolved, dealing with diverse knowing styles and choices.
Sorts Of Grid Method Multiplication Worksheet
Standard Multiplication Sheets Basic exercises focusing on multiplication tables, assisting learners develop a solid math base.
Word Problem Worksheets
Real-life scenarios integrated into issues, enhancing vital thinking and application skills.
Timed Multiplication Drills Tests designed to improve rate and accuracy, assisting in rapid mental math.
Benefits of Using Grid Method Multiplication Worksheet
The Grid Method For Multiplication 2 Digits By 2 Digits Teaching Resources
The Grid Method For Multiplication 2 Digits By 2 Digits Teaching Resources
Practice grid method multiplication with 2x1 digit calculations This KS2 Maths resource has 10 questions already written 40 questions to practice and a blank template is also available offering great
support to teachers during lesson Show more Related Searches
Lines Year 3Grid method multiplication activity suitable for children in year 3 UK This free product includes 3 differentiated sheets for children to practise using the grid method for multiplication
All problems are 2 digit x 1 digit Answers included Lines Year 3 Subjects Math Numbers
Boosted Mathematical Abilities
Constant method develops multiplication effectiveness, boosting general mathematics capabilities.
Improved Problem-Solving Talents
Word troubles in worksheets develop analytical thinking and technique application.
Self-Paced Discovering Advantages
Worksheets fit private learning rates, cultivating a comfortable and adaptable learning environment.
How to Create Engaging Grid Method Multiplication Worksheet
Integrating Visuals and Colors Vivid visuals and colors record attention, making worksheets visually appealing and involving.
Including Real-Life Circumstances
Relating multiplication to day-to-day scenarios adds importance and practicality to workouts.
Customizing Worksheets to Various Ability Degrees Personalizing worksheets based upon differing efficiency degrees makes certain comprehensive knowing. Interactive and Online Multiplication Resources
Digital Multiplication Equipment and Gamings Technology-based resources provide interactive understanding experiences, making multiplication engaging and satisfying. Interactive Sites and Apps Online
systems provide varied and available multiplication technique, supplementing conventional worksheets. Tailoring Worksheets for Numerous Understanding Styles Aesthetic Students Visual help and
diagrams help comprehension for learners inclined toward aesthetic discovering. Auditory Learners Spoken multiplication troubles or mnemonics satisfy students who understand concepts via auditory
methods. Kinesthetic Learners Hands-on tasks and manipulatives support kinesthetic students in understanding multiplication. Tips for Effective Implementation in Knowing Consistency in Practice
Routine technique reinforces multiplication abilities, advertising retention and fluency. Balancing Repetition and Variety A mix of recurring exercises and varied trouble layouts preserves interest
and comprehension. Giving Constructive Feedback Responses aids in recognizing areas of enhancement, urging ongoing development. Challenges in Multiplication Method and Solutions Motivation and
Engagement Difficulties Boring drills can lead to uninterest; innovative strategies can reignite inspiration. Getting Rid Of Concern of Math Adverse understandings around mathematics can hinder
progress; producing a positive knowing environment is essential. Impact of Grid Method Multiplication Worksheet on Academic Efficiency Research Studies and Research Searchings For Study indicates a
favorable correlation in between constant worksheet use and enhanced math performance.
Grid Method Multiplication Worksheet become functional devices, cultivating mathematical efficiency in students while suiting diverse discovering designs. From standard drills to interactive on the
internet resources, these worksheets not just enhance multiplication abilities but additionally promote essential thinking and analytical capacities.
Grid Method Multiplication Teaching Resources
Grid Method Multiplication Worksheet Pics Small Letter Worksheet
Check more of Grid Method Multiplication Worksheet below
Grid Method Of Multiplication Worksheet Teaching Resources
Grid Method Multiplying Two Digit Numbers Maths With Mum
Number Teaching Resources Number Worksheets Printable Resources On Number Cazoom Maths
Pin On Awesome Resources
Math Help How Do You Multiply Using The Grid Method Partition The Units Tens And Hundreds To
Practising The Grid Method For Short Multiplication A Worksheet Fun And Engaging PDF Worksheets
2 digit by 2 digit Multiplication with Grid Support Including
Students can use math worksheets to master a math skill through practice in a study group or for peer tutoring Use the buttons below to print open or download the PDF version of the 2 digit by 2
digit Multiplication with Grid Support Including Regrouping A math worksheet The size of the PDF file is 181300 bytes
Grid Method Multiplication Worksheets Maths Resources Twinkl
Sign Up Now to Download How can I help students practice the grid method of multiplication Practise the grid method of multiplication with this versatile bumper pack of worksheets Show more Related
Students can use math worksheets to master a math skill through practice in a study group or for peer tutoring Use the buttons below to print open or download the PDF version of the 2 digit by 2
digit Multiplication with Grid Support Including Regrouping A math worksheet The size of the PDF file is 181300 bytes
Sign Up Now to Download How can I help students practice the grid method of multiplication Practise the grid method of multiplication with this versatile bumper pack of worksheets Show more Related
Grid Method Multiplying Two Digit Numbers Maths With Mum
Math Help How Do You Multiply Using The Grid Method Partition The Units Tens And Hundreds To
Practising The Grid Method For Short Multiplication A Worksheet Fun And Engaging PDF Worksheets
Chinese Grid Method Multiplication Worksheet Times Tables Worksheets
Multiplication Grid Method Year 3 Powerpoint Jack Cook s Multiplication Worksheets
Multiplication Grid Method Year 3 Powerpoint Jack Cook s Multiplication Worksheets
2 Digit By 2 Digit Multiplication With Grid Support A Long Multiplication Worksheet
Frequently Asked Questions (Frequently Asked Questions).
Are Grid Method Multiplication Worksheet ideal for any age teams?
Yes, worksheets can be tailored to different age and ability degrees, making them versatile for various students.
How typically should trainees practice using Grid Method Multiplication Worksheet?
Constant technique is crucial. Normal sessions, ideally a few times a week, can generate substantial improvement.
Can worksheets alone boost math skills?
Worksheets are an useful tool however needs to be supplemented with diverse knowing approaches for detailed skill advancement.
Are there on the internet platforms providing free Grid Method Multiplication Worksheet?
Yes, several educational websites offer open door to a vast array of Grid Method Multiplication Worksheet.
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How To Draw The Pi Symbol
How To Draw The Pi Symbol - You can make this symbol using the key combinations explained below. Science & tech top questions what is pi? Once you release the alt key, the pi symbol should appear in
the selected cell. In this video you learn step by step. Make the top half of the 5 angular — else it looks like an s :
What is the pi symbol? Web using hex code input place the cursor on the position where you want to type the symbol. What is the value of pi? What is the symbol for pi? Web type pi in google docs to
type the pi symbol in google docs, you need to locate the insert tab. Then you can either search for pi or draw it in the box, as in the figure below. Figure 5 shows two examples of valves equipped
with positioners.
SVG > pi maths math Free SVG Image & Icon. SVG Silh
To be really simple, you can just draw two vertical lines (like an equals sign, turned 90 degrees), then draw a straight line across on top — resulting in a shape like a t with.
How to Draw the Pi Symbol Pi Step by Step Symbols Pop Culture Pi
You can continue typing with u+ input or change to abc. Web different methods of inserting a pi symbol in excel. Web 1 2 3 4 5 6 7 8 9 share no views 1.
How to Draw the Pi Symbol, Pi, Step by Step, Symbols, Pop Culture, FREE
Figure 5 shows two examples of valves equipped with positioners. Make the pi symbol (π) under windows make the symbol pi : Keep the top of the 4 open — if it closes up, it.
6 Ways to Type the Pi Symbol wikiHow
Web a positioner is symbolized by a square box on the stem of the control valve actuator. In english, π is pronounced as pie (/ p aɪ / py). Make the top half of the.
How to Draw a PI Symbol Tribal Tattoo Design Style YouTube
What is the symbol for pi? The positioner may have lines attached for motive force, instrument signals, or both. You will also have come across the symbol, π, in math, physics, and science classes.
Pi Drawing Free download on ClipArtMag
The value of pi is equal to 3.1415929. As a math student, you learned that pi is the value that is calculated by dividing the circumference of any circle by its diameter (we typically round.
6 Ways to Type the Pi Symbol wikiHow
When you click it, it will be inserted. Put a loop on the 2 so it doesn’t look like a z : To be really simple, you can just draw two vertical lines (like an.
6 Ways to Type the Pi Symbol wikiHow
Pi, in mathematics, the ratio of the circumference of a circle to its diameter. In this video you learn step by step. Keep the top of the 4 open — if it closes up, it.
6 Ways to Type the Pi Symbol wikiHow
Hold one of the option keys on your keyboard and type 03c0 to make pi symbol π. Make the pi symbol (π) under windows make the symbol pi : Alt + 9 6 0 →.
6 Ways to Type the Pi Symbol wikiHow
Web 1 2 3 4 5 6 7 8 9 share no views 1 minute ago how to draw 3d pi symbol π on paper with graphite pencil & color pencil. Web don’t slash the.
How To Draw The Pi Symbol Figure 5 shows two examples of valves equipped with positioners. 3,952 views nov 29, 2018. The positioner may have lines attached for motive force, instrument signals, or
both. Make sure to use the numeric keypad and not the numbers at the top of the keyboard. Hold one of the option keys on your keyboard and type 03c0 to make pi symbol π.
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Significant Figures
Significant Figures
Chapter 1R
Audio created by Google NotebookLM. Note: I have not verified the accuracy of the audio or transcript. Download the transcript here (created by Restream).
Your browser does not support the audio tag.
Read more about Significant Figures on Wikipedia.
Counting Sig. Figs.
Note: Given inexact numbers should contain all certain digits and the first uncertain digit.
1. Count left-to-right when counting significant figures.
2. All non-zero digits of a provided number are significant.
1 significant figure
2 significant figures
3 significant figures
3. Zeroes between two other significant digits are significant.
3 significant figures
4. Zeroes to the right of a non-zero number, and also to the right of a decimal place, are significant
4 significant figures
5. Zeroes that are placeholders are not significant.
2 significant figures
Ambiguous (could be 2, 3, 4, or 5). For the purpose of this class, 2 significant figures.
Use of a decimal place at the end would make the trailing zeroes significant.
5 significant figures
Use of scientific notation can also remove the ambiguity.
\[1.3\times 10^4\]
2 significant figures
\[1.30\times 10^{4}\]
3 significant figures
\[1.300\times 10^{4}\]
4 significant figures
\[1.300~0\times 10^{4}\]
5 significant figures
6. Exact numbers (those obtained by counting) have an infinite number of significant figures. Fractions can be exact. Physical constants such as the molar gas constant can also be exact if derived
from other exact numbers. See NIST to determine if a physical constant is exact or inexact (i.e. has a non-zero uncertainty).
If this number was obtained by counting (not measuring), it is exact and has an infinite number of significant figures. For example, to say that there is 60 seconds in 1 minute would make the “60”
exact. If 60 s was the result of a measurement, it would have 1 significant figure.
If one were to count 30 lemons, the number “30” would be an exact number with an infinite number of significant figures.
7. Some conversion factors are exact while others are inexact. For example, 1 inch is defined as being exactly 2.54 cm. Therefore, both values in the following conversion factor, 1 in = 2.54 cm, are
exact. However, 1 gallon (US) is approximately equal to 3.785412 L (inexact). The quantity given for L is inexact and would have 7 significant figures.
8. A mathematical or physical constants has significant figures to its known digits. For example, as of March 2024, π is known to 102 trillion digits, each of which are significant. Constants such
as speed of light (c), gas constant (R), etc. also fall into this category. Note: Using a rounded off physical constant (e.g. 3.00×10^8 m s^–1 instead of 299 792 458 m s^–1 for speed of light)
will limit the number of significant figures for that constant. See NIST to determine if a physical constant is exact or inexact (i.e. has a non-zero uncertainty).
\[c = 299~792~458~\mathrm{m~s^{-1}}\]
This value for the speed of light is exact and has an inifinite number of significant figures.
\[c = 3.00\times 10^{8}~\mathrm{m~s^{-1}}\]
This value for the speed of light has 3 significant figures.
\[R = 8.314~462~618~153~24~\mathrm{J~mol^{-1}~K^{-1}}\]
This value for the molar gas constant is exact and has an infinite number of significant figures.
\[R = 8.314~\mathrm{J~mol^{-1}~K^{-1}}\]
This value for the molar gas constant is inexact and has 4 significant figures.
Significant Figures in Calculations
1. When adding or subtracting numbers, the result contains no significant digits beyond the place of the last significant digit of any datum.
\[\begin{align*} 3.24 + 1.9 + 12.482 &= 17.\bar{6}22 \\[1.5ex] &= 17.6 \\[4ex] 5.421 - 10.138 + 3.41 &= -1.3\bar{0}7 \\[1.5ex] &= -1.31 \\[4ex] 346 - 343.4 &= \bar{2}.6 \\[1.5ex] &= 3 \\[4ex] 25 -
10.1 &= 1\bar{4}.9 \\[1.5ex] &= 15 \\[4ex] 10. + 16.3 &= 2\bar{6}.3 \\[1.5ex] &= 26 \\[4ex] 10.0 + 16.3 &= 26.3 \end{align*}\]
For numbers that clearly have an ambiguous number of significant figures, assume the zeroes to be insignificant (for the purpose of this class).
\[\begin{align*} 210~000 + 61~435 &= 2\bar{7}1~435 \\[1.5ex] &= 270~000 \end{align*}\]
Here, 210 000 has an ambiguous number of significant figures. The last non-zero digit is located in the ten thousands place. The result should be rounded to the ten thousands place.
Here are a few more examples.
\[\begin{align*} 23~100 + 32 &= 23~\bar{13}32 \\[1.5ex] &= 23~100 \\[4ex] 890 + 12 &= 9\bar{0}2 \\[1.5ex] &= 900 \\[4ex] 312 + 300~000 &= \bar{3}00~312 \\[1.5ex] &= 300~000 \\[4ex] 10 + 16.3 &= \bar
{2}6.3 \\[1.5ex] &= 30 \end{align*}\]
2. In multiplication or division, the number of significant figures in the answer is determined by the value with the fewest significant figures.
\[\begin{align*} 3.24 \times 812.3 &= 2~6\bar{3}1.852 \\[1.5ex] &= 2.63\times 10^{3}\\[4ex] 1.502 \left ( \dfrac{4.90}{2.11} \right ) &= 1.502 \left ( 2.3\bar{2}2\right ) \\[1.5ex] &= 3.4\bar{8}80 \\
[1.5ex] &= 3.49 \\[4ex] \left ( 346 - 343.4 \right ) / 8.4 &= \left ( \bar{2}.6 \right ) 8.4 \\[1.5ex] &= 0.3095 \\[1.5ex] &= 0.3 \end{align*}\]
3. As per the textbook, when a number is rounded off, the last digit retained is increased by one (rounded up) only if the following digit is 5 or greater. NOTE: The round-half-to-even rule is
followed by NIST, ANSI, ASTM, etc. where a number only gets rounded up on a 5 if the resulting number was an even number.
Each number below is rounded to 3 significant figures using the rule provided by the textbook.
\[\begin{align*} 12.\bar{6}96 &\rightarrow 12.7 \\[1.5ex] 18.\bar{3}49 &\rightarrow 18.3 \\[1.5ex] 14.\bar{9}99 &\rightarrow 15.0 \\[1.5ex] 14.\bar{3}5 &\rightarrow 14.4 \\[1.5ex] 1.1\bar{2}51 &\
rightarrow 1.13 \end{align*}\]
The following examples round the given numbers to 3 significant figures using the round-half-to-even rule.
\[\begin{align*} 14.\bar{3}5 &\rightarrow 14.4 \\[1.5ex] 1.1\bar{2}51 &\rightarrow 1.12 \end{align*}\]
4. A rounded value should be obtained in one step by direct rounding of the most precise value available and not in two or more successive roundings. For example: 89 490 rounded to the nearest 1 000
is at once 89 000; it would be incorrect to round first to the nearest 100, giving 89 500 and then to the nearest 1 000, giving 90 000.
5. In a multi-step calculation, only round the final value. Determine the number of significant figures in the final result by considering each step in the calculation. In intermediate steps, write
the number to the proper number of significant figures and keep at least one additional digit.
\[\begin{align*} (2.349~4 + 1.345) \times 1.2 &= 3.69\bar{4}~4 \times 1.2 \\[1.5ex] &= 4.\bar{4}3 \\[1.5ex] &= 4.4 \\[4ex] (2.349~4 \times 1.345) + 1.2 &= 3.15\bar{9}~9 + 1.2 \\[1.5ex] &= 4.\bar{3}5
\\[1.5ex] &= 4.4 \end{align*}\]
6. Digits in logarithms, ln(x) or log[10](x), are significant through the n-th place after the decimal when x has n significant figures. A quantity resulting from a logarithmic transformation is
\[\ln(3.46~\mathrm{kPa}) = 1.241\]
3.46 has 3 significant figures. The result should have three places after the decimal. Note that the resulting number is dimensionless (has no unit).
\[\log(3.000\times 10^4) = 4.477~1\]
3.000 × 10^4 has 4 significant figures. The result should have 4 places after the decimal.
7. Significant digits as a result of exponentials and antilogarithms, e^x or 10^x, is equal to the place of the last significant digit in x after the decimal. A number resulting from an antilog
transformation will have dimensions if the original logarithmic value was derived from a quantity with units.
\[e^{1.241} = 3.4\bar{5}9~07 = 3.46\]
Since 1.241 has 3 decimal places, the final answer should have 3 significant figures.
\[10^{4.4771} = 29~9\bar{9}8.531~811~9\ldots = 30~000 = 3.000\times 10^{4}\]
Since 4.477 1 has 4 decimal places, the final answer should have 4 significant figures. | {"url":"https://dornshuld.com/chem1/notes/ch01r-significant-figures.html","timestamp":"2024-11-14T13:44:27Z","content_type":"application/xhtml+xml","content_length":"49786","record_id":"<urn:uuid:1c2006e8-2368-4832-9a2d-54da49c2e183>","cc-path":"CC-MAIN-2024-46/segments/1730477028657.76/warc/CC-MAIN-20241114130448-20241114160448-00401.warc.gz"} |
11.1 Facts About the Chi-Square Distribution
The notation for the chi-square distribution is
$χ∼ χ df 2 χ∼ χ df 2$
where df = degrees of freedom, which depends on how chi-square is being used. If you want to practice calculating chi-square probabilities then use df = n – 1. The degrees of freedom for the three
major uses are calculated differently.
For the χ^2 distribution, the population mean is μ = df, and the population standard deviation is $σ= 2(df) σ= 2(df)$.
The random variable is shown as χ^2, but it may be any uppercase letter.
The random variable for a chi-square distribution with k degrees of freedom is the sum of k independent, squared standard normal variables is
χ^2 = (Z[1])^2 + (Z[2])^2 + ... + (Z[k])^2, where the following are true:
• The curve is nonsymmetrical and skewed to the right.
• There is a different chi-square curve for each df.
• The test statistic for any test is always greater than or equal to zero.
• When df > 90, the chi-square curve approximates the normal distribution. For X ~ $χ 1,000 2 χ 1,000$, the mean, μ = df = 1,000 and the standard deviation, σ = $2(1,000) 2(1,000)$ = 44.7.
Therefore, X ~ N(1,000, 44.7), approximately.
• The mean, μ, is located just to the right of the peak. | {"url":"https://texasgateway.org/resource/111-facts-about-chi-square-distribution?book=79081&binder_id=78266","timestamp":"2024-11-11T08:34:30Z","content_type":"text/html","content_length":"42256","record_id":"<urn:uuid:8aacae7d-d980-433b-9f04-7a1df0cf69e6>","cc-path":"CC-MAIN-2024-46/segments/1730477028220.42/warc/CC-MAIN-20241111060327-20241111090327-00650.warc.gz"} |
Maths Past Questions and Answers | PastQuestions.com.ng
Maths Past Questions and Answers
Maths Past Questions and Answers – Are you an intending candidate and you are wondering how you will make it in Mathematics in general? and you would love to participate in this year’s admission
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Maths Past Questions and Answers Pattern
Normally, the Maths Past Questions and Answers are patterned in three ways. there are the objectives and theory, and for some subjects, there is the practical aspect. We have made it very easy for
you. we bring all the questions for many years and put them together but we indicate the specific years of their occurrence. We provide the correct answers to save you time. All you need to do is
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If you would not like to be disappointed during the Jamb examination, then it’s high time you buckled down for the examination. You need this past question for the following under listed reasons;
Getting the Maths Past Questions and Answers from us, and taking time to study the content will expose you to the questions that have been asked years back as most of those questions and answers get
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1. It, however, provides useful information about the types of questions to expect and helps you prevent unpleasant surprises.
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Maths Past Questions and Answers Sample
In other to know that we are giving out the original Maths Past Questions and Answers, we have decided to give you some free samples as proof of what we said earlier. Jamb Result
Question 1
Differentiate (2x+5)2(x−4) with respect to x.
A) 4(2x+5)(x−4)
B) 4(2x+5)(4x−3)
C) (2x+5)(2x−13)
D) (2x+5)(6x−11)
The correct answer is D.
To differentiate (2x+5)2(x−4), you first need to know that it is a product function.
Using the product rule you have dydx=udvdx+vdudx
Let (2x+5)2 be u and (x−4) be v.
To find du we use the chain rule which is dudx=dudw×dwdx
We say let (2x+5) be w
We then have a new function u=w2
So dudx=2×2w which equals 4w and w was (2x+5)
Substituting everything into the product rule we have:
Question 2
Find the area bounded by the curves y=4−x2 and y=2x+1
A) 2013 sq. units
B) 2023 sq. units
C) 1023 sq. units
D) 1013 sq. units
The correct answer is C.
y=4−x2 and y=2x+1
Thus, x=−3 or x=1.
Integrating x2+2x−3 from (−3 to 1) concerning x will give 323=1023
Question 3
Find the rate of change of the volume, V of a sphere concerning its radius, r when r=1.
A) 12π
B) 4π
C) 24π
D) 8π
The correct answer is B.
The volume of the sphere, V=43×πr3
Rate of change of V=dvdr
Thus if, V=43×πr3,
At r=1, Rate =4×π×1=4π
Question 4
If y=xsinx, find dydx when x=π2.
A) −π2
B) −1
C) 1
D) π2
The correct answer is C.
At x=π2, =sinπ2+π2cosπ2
Question 5
Find the dimensions of a rectangle of the greatest area which has a fixed perimeter p.
A) square of sides p
B) square of sides 2p
C) square of sides p2
D) square of sides p4
The correct answer is D.
Let the rectangle be a square of sides p4.
So that perimeter of square =4p
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7 Tips to Prepare for Maths Exams
1. Don’t make reading your hobby: A lot of people put reading as a hobby in their CV, they might be right because they have finished schooling. But “You” are still schooling, so reading should be a
top priority, not a hobby. Read far and wide to enhance your level of aptitude
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mastery of the scope of the exams you are expecting.
3. Attend Extramural Classes: Register and attend extramural classes at your location. This class will help you refresh your memory and boost your classroom understanding and discoveries of new
4. Sleep when you feel like: When you are preparing for any exams, sleeping is very important because it helps in the consolidation of memory. Caution: Only sleep when you feel like it and don’t
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AIcrowd | RL-Taxi | Challenges
π ΅οΈ Introduction
In this problem we have a a taxi driver, who serves three cities A, B and C. On a regular workday, the taxi driver can find a new ride by choosing one of the following actions:
1. Cruise the streets looking for a passenger.
2. Go to the nearest taxi stand and wait in line.
3. Wait for a call from the dispatcher (this is not possible in town B because of poor reception).
For a given town and a given action, there is a probability that the next trip will go to each of the towns A, B and C and a corresponding reward in monetary units associated with each such trip.
This reward represents the income from the trip after all necessary expenses have been deducted. Please refer to the table below for the rewards and transition probabilities.
• p^k[ij] is the probability of getting a ride to town j, by choosing an action k while the driver was in town i.
• r^k[ij] is the immediate reward of getting a ride to town j, by choosing an action k while the driver was in town i.
1. Implement the DP algorithm. Find the pseudocode below. Let S be the state space and A the action space.
2. Tabulate the optimal policy and optimal value for each state in each round for N = 10.
3. Consider a policy that always forces the driver to go to the nearest taxi stand, irrespective ofthe state. Is it optimal? Justify your answer.
You will be writing your solutions & making a submission through a notebook. You can follow the instructions in the starter notebook.
π Ύ Dataset
π Files
Under the Resources section you will find data files that contains parameters for the environment for this problem.
π Submission
Submissions will be made through a notebook following the instructions in the starter notebook.
π ± Contact | {"url":"https://www.aicrowd.com/challenges/rliitm-1/problems/rl-taxi","timestamp":"2024-11-12T12:21:09Z","content_type":"text/html","content_length":"212757","record_id":"<urn:uuid:f36e8c0b-1ea0-40fd-991a-9cab4aa31f70>","cc-path":"CC-MAIN-2024-46/segments/1730477028273.45/warc/CC-MAIN-20241112113320-20241112143320-00157.warc.gz"} |
Understanding Digital Signatures: More Than Just a Hash
Digital signatures are a cornerstone of modern security practices, ensuring data integrity and authentication in various online communications. But there's often confusion between the terms "digital
signature", "hash", and "digest". Let's delve deep into understanding these terms and their roles.
What is a Digital Signature?
At its core, a digital signature is a mechanism used to verify the authenticity and integrity of a message, software, or digital document. It's like an electronic stamp, proving that the content
hasn't been altered since it was signed and verifying the identity of the signer.
Properties of a Digital Signature:
• Authentication: Validates the sender's identity.
• Data Integrity: Ensures that the content hasn't changed since being signed.
• Non-repudiation: Signers cannot deny having signed the content.
Hash vs. Signature
A hash is a function that converts an input (often called a "message") into a fixed-length string of bytes, which appears random. This output is commonly referred to as the hash value or digest.
On the other hand, a digital signature involves more than just output encodings and hashing. It uses a signing algorithm, which often employs a hash function as one of its steps. But, importantly, it
also incorporates elements like symmetric keys to create a unique signature, even with the same message input.
Understanding the Digest
A digest is the encoded output of a hash function. It's the "end result" you get after passing your data through the hash function. The term "digest" is often used interchangeably with "hash value",
emphasising the output's nature.
A hexadecimal output is precisely what is referred to as the digest, it's just an encoded output for the chosen hash. It could as easily be encoded as binary, base64, whatever encoding - it is still
an output of the same hash.
So there's the twist: looking only at a digest, it's impossible to determine whether it resulted from a hash function alone or was part of a digital signature process. Both can produce
similar-looking outputs, but their origins are different.
Signature Algorithms: The Heart of Digital Signatures
Every digital signature is rooted in a particular signing algorithm. This algorithm determines how the signature is produced and, consequently, how it will be verified. Here are a few examples:
• HMAC-SHA512: This combines the HMAC (Hash-Based Message Authentication Code) method with the SHA-512 hash function.
• ECDSA-MD5: Uses Elliptic Curve Digital Signature Algorithm with the MD5 hash function.
• RSA-SHA3-512: Combines the RSA (Rivest–Shamir–Adleman) algorithm with the SHA3-512 hash function.
To put it simply:
• HMAC-SHA512, ECDSA-MD5, and RSA-SHA3-512 are signatures.
• HMAC-SHA256, ECDSA-SHA256, and RSA-SHA256 are also signatures.
• SHA256, on its own, is just a hash.
• Digests can be either hash or signatures, the determining factor is the method to reproduce and therefore verify the digest is one or the other.
Here's how you can create a HMAC-SHA512 signature in Python and then verifying it in JavaScript.
Python (Sender):
import hmac
import hashlib
import base64
def generate_hmac_sha512_signature(secret_key, message):
signature = hmac.new(secret_key.encode(), message.encode(), hashlib.sha512).digest()
return base64.b64encode(signature).decode()
secret_key = "supersecretkey"
message = "Hello, World!"
signature = generate_hmac_sha512_signature(secret_key, message)
This code generates an HMAC-SHA512 signature using Python's hashlib library, JavaScript (Receiver) verifies the HMAC-SHA512 signature using Node.js's crypto library:
const crypto = require('crypto');
function verifyHMACSHA512Signature(secretKey, message, signature) {
const hmac = crypto.createHmac('sha512', secretKey);
const computedSignature = hmac.digest('base64');
return computedSignature === signature;
const secretKey = "supersecretkey";
const message = "Hello, World!";
const receivedSignature = "..."; // This should be the output from the Python script
if (verifyHMACSHA512Signature(secretKey, message, receivedSignature)) {
console.log("Signature is valid!");
} else {
console.log("Signature is NOT valid!");
For ECDSA-SHA256 signing in Python, we can use the ecdsa library in a very similar way. The signature can then be verified in JavaScript using the elliptic library in an ECMAScript module.
import ecdsa
import base64
def generate_ecdsa_sha256_signature(secret_key, message):
sk = ecdsa.SigningKey.from_string(bytes.fromhex(secret_key), curve=ecdsa.NIST256p)
signature = sk.sign(message.encode())
return base64.b64encode(signature).decode()
secret_key = "your_private_key_in_hex_format"
message = "Hello, World!"
signature = generate_ecdsa_sha256_signature(secret_key, message)
The JavaScript verifier is also not too different:
import { ec } from 'elliptic';
function verifyECDSASHA256Signature(publicKey, message, signature) {
const ecInstance = new ec('p256');
const key = ecInstance.keyFromPublic(publicKey, 'hex');
const isValid = key.verify(message, Buffer.from(signature, 'base64').toString('hex'));
return isValid;
const publicKey = 'your_public_key_in_hex_format';
const message = 'Hello, World!';
const receivedSignature = '...'; // This should be the output from the Python script
if (verifyECDSASHA256Signature(publicKey, message, receivedSignature)) {
console.log('Signature is valid!');
} else {
console.log('Signature is NOT valid!');
Make sure you have the necessary libraries installed, e.g.:
• Python: pip install ecdsa
• JavaScript: npm install elliptic
Hash Verification:
When you hash data, the outcome is a fixed-length string of characters, regardless of the input's size. To verify a hash, you take the original message data and run it through the same hashing
algorithm again. If the resultant digest matches the previously produced hash, then the data hasn't been tampered with. Essentially, you're reproducing the hash digest with the message data on both
ends to ensure they match.
Symmetric Signature Verification (HMAC):
HMAC (Hash-based Message Authentication Code) involves a hash function and a symmetric secret key. When sending a message, you encipher the data using this symmetric key, creating an HMAC signature
digest. To verify, the receiver, who also has the symmetric key, will reproduce the HMAC signature from the received message data. If the digests match, it verifies the message's integrity and
authenticates its origin, since only someone with the shared secret key could produce the same signature.
Asymmetric Signature Verification (ECDSA):
Elliptic Curve Digital Signature Algorithm (ECDSA) involves an asymmetric key pair: a private key and a corresponding public key. The sender uses the private key to create a signature for the message
data. The receiver, or any verifier, uses the sender's public key to verify the signature. The beauty of this method is that verification ensures both the data's integrity and the sender's
authenticity. Only the holder of the private key could've produced a signature that the public key can verify, yet the private key itself isn't exposed during this process.
Verification assures Sender Authentication
Verification is paramount for sender authentication. While a simple hash can guarantee data integrity (i.e., the data hasn't changed), it doesn't confirm who sent the data. HMAC, with its symmetric
key, offers an extra layer of authentication. However, ECDSA and other asymmetric methods add an even stronger assurance. Since only the private key holder can sign the message in a way that the
corresponding public key can verify, receivers can trust not only the message's content but also its source.
Attacks and Weakness
I can't talk about secrets and authentication without at least discussing the threat vectors! Here are a mix of known and potential attacks to think about:
• Collision (Birthday) Attacks: Two different sets of data produce the same hash digest. MD5 and SHA-1 are known to be vulnerable to this attack. While SHA-256 is considered secure, its resistance
depends on the continuing evolution of computational power and cryptanalysis techniques, and it is unclear if cryptocurrency mining GPUs or ASIC miners have been used for this kind of research..
• Pre-image (rainbow tables / dictionary) and Second Pre-image Attacks: Finding a message that hashes to a specific target hash (pre-image) or a different message with the same hash as the original
message (second pre-image).
• Length Extension Attacks: Exploiting the mathematical properties of hash functions to append additional data to the message. Particularly possible when not using a HMAC. A length extension attack
exploits properties of certain hash functions where, given a hash and length of the original data, new data can be added.
• Potential Attack Vectors: Quantum computers pose a theoretical threat to hash functions due to their potential to perform complex calculations faster than classical computers.
• Brute Force Attacks: Attempting all possible keys until the correct one is found.
• Chosen-plaintext Attack: Attackers choose specific plaintexts to be encrypted and analyse the ciphertexts or public key to gather information about the private key. This is a weakness of block
cipher modes
• Known Plaintext Attacks: Using known parts of the plaintext and ciphertext to derive the key. This happened with Debian's (and subsequent others) factorable RSA and DSA public keys, and before
that there was the SHAmbles attack.
• Inherent Trust Attacks: Relies on both parties having the secret key. If the key is exposed, data integrity and authenticity are compromised (never was assured).
• Man-in-the-Middle Attacks: An attacker intercepts and potentially alters the communication between two parties without them knowing.
• Replay Attacks: An adversary captures legitimate encrypted data and later resends it, aiming to cause unauthorized actions or reveal information about the key.
• Side-channel Attacks: Attackers gain information from the physical system performing the encryption, like power consumption or acoustic emissions.
• Private Key Derivation: Theoretically, if an attacker has enough data and computational power, they might derive the private key from public components, although currently, this is practically
impossible for strong asymmetric cryptography.
• Forward Secrecy Violation: If reused keys aren't regularly changed, an attacker who gets the long-term private key can decrypt past encrypted data, or forge signatures, with the same key.
In Conclusion
While hashes and digital signatures may seem similar on the surface, they serve different purposes in the realm of security. A hash ensures data integrity, while a digital signature ensures both data
integrity and sender authentication.
Understanding this difference is crucial for verification.
Verification is the only way you gain any of the benefits at all. | {"url":"https://www.langton.cloud/understanding-digital-signatures-more-than-just-a-hash/","timestamp":"2024-11-04T21:48:20Z","content_type":"text/html","content_length":"32937","record_id":"<urn:uuid:d6f09297-9c70-44a9-b544-f3a17e2078dc>","cc-path":"CC-MAIN-2024-46/segments/1730477027861.16/warc/CC-MAIN-20241104194528-20241104224528-00606.warc.gz"} |
Видеотека: T. Jung, $p$-Laplacian boundary value problem with jumping nonlinearities
Аннотация: We investigate multiplicity of solutions for one dimensional $p$-Laplacian Dirichlet boundary value problem with jumping nonlinearites. We obtain three theorems: The first one is that
there exists exactly one solution when nonlinearities cross no eigenvalue. The second one is that there exist exactly two solutions, exactly one solutions and no solution depending on the source term
when nonlinearities cross one first eigenvalue. The third one is that there exist at least three solutions, exactly one solutions and no solution depending on the source term when nonlinearities
cross the first and second eigenvalues. We obtain the first theorem and the second one by eigenvalues and the corresponding normalized eigenfunctions of the $p$-Laplacian Dirichlet eigenvalue
problem, and the contraction mapping principle on $p$-Lebesgue space (when $p \geqslant 2$). We obtain the third result by Leray-Schauder degree theory.
This is a joint work with Q-Heung Choi (Inha University, Incheon, South Korea).
Язык доклада: английский | {"url":"https://m.mathnet.ru/php/presentation.phtml?option_lang=rus&presentid=24787","timestamp":"2024-11-07T13:49:09Z","content_type":"text/html","content_length":"8891","record_id":"<urn:uuid:636b88d1-6b34-4cb6-bc9b-361cf675726d>","cc-path":"CC-MAIN-2024-46/segments/1730477027999.92/warc/CC-MAIN-20241107114930-20241107144930-00211.warc.gz"} |
Liouville's number, the easiest transcendental and its clones (corrected reupload) | Video Summary and Q&A | Glasp
Liouville's number, the easiest transcendental and its clones (corrected reupload) | Summary and Q&A
Liouville's number, the easiest transcendental and its clones (corrected reupload)
Liouville's number is a transcendental number with a unique decimal expansion pattern, and it can be used to create a clone of the real numbers within the real numbers.
Key Insights
• 👍 Liouville's number is transcendental and its transcendence can be proven through an examination of the digit patterns of its squared truncations.
• #️⃣ The decimal expansion of Liouville's number allows for the creation of a clone of the real numbers within the real numbers, consisting entirely of transcendental numbers.
• #️⃣ The clone of the real numbers based on Liouville's number has measure zero, indicating that it takes up no space within the real numbers.
Welcome to another Mathologer video. Liouville's number the monster up there consists of infinitely many isolated islands of 1s at the 1! th, 2! th, 3! th, etc. digits with exploding gaps of zeros
between them. As I promised you at the end of the last video, today's mission is to show you a nice visual way of seeing that this number is transcendent... Read More
Questions & Answers
Q: What is Liouville's number and how does its decimal expansion pattern make it transcendental?
Liouville's number is a transcendental number with a decimal expansion consisting of isolated islands of 1s and exploding gaps of zeros between them. This unique pattern of digits, with the
increasing gaps between the 1s, contributes to its transcendence.
Q: How is the transcendence of Liouville's number proven?
The proof of Liouville's number being transcendental involves examining the digit patterns of the squared truncations of the number. It is demonstrated that, for each power of Liouville's number,
there is a certain point from which all the digits of the truncations are correct, indicating its transcendence.
Q: Can Liouville's number be used to create a clone of the real numbers within the real numbers?
Yes, using Liouville's number as a template, it is possible to create a clone of the real numbers consisting entirely of transcendental numbers. This clone has the same cardinality as the set of real
numbers but has measure zero, meaning it takes up no space within the real numbers.
Q: How is the decimal expansion of Liouville's number related to the creation of the clone of the real numbers?
The decimal expansion of Liouville's number, with isolated islands of 1s at specific digit locations, is used to create a clone of the real numbers by replacing the 1s with other digits. As long as
infinitely many of these replacement digits are nonzero, the resulting numbers in the clone will also be transcendental.
Summary & Key Takeaways
• Liouville's number is a transcendental number with a decimal expansion consisting of isolated islands of 1s and exploding gaps of zeros between them.
• The proof of Liouville's number being transcendental is accessible to those with some exposure to real analysis, and it involves examining the digit patterns of the squared truncations of the
• Using Liouville's number as a template, it is also possible to create a clone of the real numbers consisting entirely of transcendental numbers, which has measure zero.
Explore More Summaries from Mathologer 📚 | {"url":"https://glasp.co/youtube/p/liouville-s-number-the-easiest-transcendental-and-its-clones-corrected-reupload","timestamp":"2024-11-09T01:36:09Z","content_type":"text/html","content_length":"358425","record_id":"<urn:uuid:ad490e2f-9cd9-49d5-a363-3f406fe07bbd>","cc-path":"CC-MAIN-2024-46/segments/1730477028106.80/warc/CC-MAIN-20241108231327-20241109021327-00829.warc.gz"} |
Multiplication Chart 1 23 2024 - Multiplication Chart Printable
Multiplication Chart 1 23
Multiplication Chart 1 23 – If you are looking for a fun way to teach your child the multiplication facts, you can get a blank Multiplication Chart. This can allow your kid to fill the details by
themselves. You can get empty multiplication graphs for a variety of product or service varieties, which includes 1-9, 10-12, and 15 products. If you want to make your chart more exciting, you can
add a Game to it. Below are a few tips to get the kid began: Multiplication Chart 1 23.
Multiplication Graphs
You can utilize multiplication maps in your child’s university student binder to assist them to memorize arithmetic information. Although children can remember their math concepts details normally,
it takes many others time to accomplish this. Multiplication maps are a good way to reinforce their boost and learning their self confidence. As well as being educative, these graphs could be
laminated for added sturdiness. Allow me to share some valuable methods to use multiplication maps. You can also have a look at these websites for helpful multiplication fact sources.
This lesson includes the basics of your multiplication dinner table. As well as studying the rules for multiplying, students will recognize the very idea of factors and patterning. Students will be
able to recall basic facts like five times four, by understanding how the factors work. They may also be able to utilize your property of one and zero to eliminate more complicated items. Students
should be able to recognize patterns in multiplication chart 1, by the end of the lesson.
Besides the common multiplication graph, students may need to produce a graph or chart with increased factors or fewer aspects. To produce a multiplication graph or chart with more aspects, students
need to produce 12 furniture, each with a dozen rows and about three columns. All 12 dining tables should in shape on a single page of document. Lines should be drawn using a ruler. Graph papers is
perfect for this undertaking. If graph paper is not an option, students can use spreadsheet programs to make their own tables.
Online game concepts
Whether you are instructing a newbie multiplication session or working on the mastery of your multiplication desk, you are able to come up with entertaining and engaging online game concepts for
Multiplication Graph 1. Several enjoyable suggestions are the following. This game necessitates the pupils to remain work and pairs on a single issue. Then, they are going to all hold up their
greeting cards and go over the solution for the min. They win if they get it right!
When you’re teaching kids about multiplication, among the finest resources it is possible to allow them to have is really a computer multiplication graph or chart. These printable sheets appear in a
number of models and may be published using one page or numerous. Children can discover their multiplication specifics by copying them from the memorizing and chart them. A multiplication chart may
help for several factors, from supporting them find out their arithmetic facts to educating them using a calculator.
Gallery of Multiplication Chart 1 23
Multiplication Table Poster For Kids Educational Times Table Chart
Multiplication Table Poster Chart Laminated For Kids And Math Classroom
MULTIPLICATION TABLE Multiplication Table Multiplication Table
Leave a Comment | {"url":"https://www.multiplicationchartprintable.com/multiplication-chart-1-23-2/","timestamp":"2024-11-02T12:46:47Z","content_type":"text/html","content_length":"53153","record_id":"<urn:uuid:c9413cdc-7a05-4fb4-8e3d-0d63910601b1>","cc-path":"CC-MAIN-2024-46/segments/1730477027710.33/warc/CC-MAIN-20241102102832-20241102132832-00537.warc.gz"} |
Check if a Char Variable is a Space Character
Assume that c is a char variable that has been declared and already given a value . Write an expression whose value is true if and only if c is a space character .
LANGUAGE: C++
Assume that c is a char variable that has been declared and already given a value . Write an expression whose value is true if and only if c is a space character. | {"url":"https://matthew.maennche.com/2013/12/assume-that-c-is-a-char-variable-that-has-been-declared-and-already-given-a-value-write-an-expression-whose-value-is-true-if-and-only-if-c-is-a-space-character/","timestamp":"2024-11-08T21:08:29Z","content_type":"text/html","content_length":"90842","record_id":"<urn:uuid:327a55ec-9fc4-4347-9cc6-9c66a9445eca>","cc-path":"CC-MAIN-2024-46/segments/1730477028079.98/warc/CC-MAIN-20241108200128-20241108230128-00473.warc.gz"} |
Problem-Solving Strategies
In those note, I enumerate certain strategies and give advice on approaching the midterm.
Understand Your Objective
You may think, most likely subconsciously, that your objective on an exam should be to solve all problems. Wrong. Your objective should be to get as many point as you can. As soon as you understand
this, your approach to solving problems should change drastically.
Instead of getting stuck on any given problem and trying to tackle it until you solve it — you just distribute your time evenly over all problems. Imagine you have 50 minutes and 6 problems (fyi, I
have no idea how many problems you will have on an actual exam). I’d leave 10-20% of time for review, in this case 8 minutes, and then distribute the rest evenly: 7 min per each problem (). Start
working on problem 1 and do as much as you can in 7 min. After 7 minutes, it doesn’t matter whether you solved the problem or not. Move on to problem 2. It is hard to do so, you may feel it’s wrong
to leave the problem unsolved, you may think oh, if I just spend a few more minutes, I may solve the problem. But which outcome is better, have 2 fully solved problems and receive 0 marks on the
other 4, or have partial solution to all 6 problems? Aesthetically, I could agree the former is better. Point-wise, it’s very likely you’ll have a greater score in the latter outcome.
Translate the Problem from English to Math
Consider the following problem:
In a reaction , the is consumed at a rate of . What is the rate at which is produced?
Something I observed many of you do when solving problems like this, is write "" in your notes and then try to argue. But then it’s very easy to confuse what this value represents? You may think it’s
the reaction rate, and then say okay, the reaction rate expressed through is , therefore the answer is , which would be incorrect.
Imagine if I gave you the problem set up as follows:
In a reaction , the value of is equal to . Find .
I’m confident all of you can solve the problem in this formulation. As long as you understand the following identity:
All you need to do is simply plug in the value for !
Before you start solving any problem, your task is to translate it from English into symbolic representation. Do not allow yourself to write any values without labels. Read the problem, find every
piece of data presented in the problem, convert it into symbolic form and write it down in your solutions.
Double Check your Interpretation of the Problem
Consider the following problem:
is a reactant which is consumed in a process obeying first-order kinetics with the rate constant . Find the time in which the concentration of decreases to one third of its starting
Similar problem was in one of the worksheets, and a common approach looked like this:
Then you may try to plug in the values and find that you get a negative answer (because the log of a concentration smaller than 1 is negative). But no matter how long you look at your work, you will
not find any issues with it. Because there are no algebraic mistakes. The mistake is in the interpretation of the problem. The correct equation should be:
But I’d argue that you shouldn’t skip preceding steps. The problem, written in English, states neither (3.1) nor (3.2) directly. What it says is:
• is consumed in the first-order process. Therefore, we can say that .
• We need to find when A decreases to one-third of its starting concentration. Thus, we seek . You can obtain (3.2) by plugging the second condition into the equation from the first condition. If
you skip this step and you try to write equation 3.2 directly, you may make a mistake and write 3.1 and it’ll be very difficult to find where you made a mistake.
Do not guess. Formulate complete logical statements.
Consider the following problem:
The Haber process is an exothermic reaction. What will happen to the equilibrium constant if we increase the temperature?
The correct answer can be expressed with a single word, so it may be tempting to immediately try and say that word: decrease. The problem is, you probably won’t have a lot of confidence in your
answer. You might as well say increase. How do you know if you’re right? Can you tell whether you’re right by looking at the answer? No.
What you should do is to formulate complete logical statements like this:
The reaction is exothermic, which means that heat/energy is released during the forward reaction. If we increase the temperature, we essentially add energy/heat to the system. The system responds
to our efforts by trying to minimize the influence of our efforts (Le-Chatelier principle). Therefore, the system will try to decrease the amount of energy. It can do so through the reverse
reaction (if the forward reaction releases energy the reverse must absorb it). Therefore, the system will prioritize the reverse reaction and so the amount of reactants will increase. Because
reactants are in the denominator of the equilibrium constant, the value of the equilibrium constant will decrease.
Is it longer? Yes. But it’s also more foolproof. You can check whether you have the correct answer or not simply by evaluating whether each individual sentence is true and then whether the logical
connections (because/therefore) are used correctly. With practice (and time) you may be able to start doing this reasoning in your head or even subconsciously. But in the beginning, you should write
it out explicitly. Not only because it’ll increase the chance that you find the correct answer, but also because you can get partial credit for your solution.
Imagine if you make a mistake and interpret “exothermic” as heat is absorbed during the forward reaction. Your final answer will be that the value of will increase, which will be a wrong answer. And
if you write just that answer, you will get 0 points. However, if you write the whole paragraph above, it’ll be easy to see that you have the correct reasoning and you made a minor mistake in
interpretation of the terminology. So you will get partial credit.
Do not underestimate psychology
If you look at mistakes like above (confusing whether heat is released or absorbed in exothermic reactions) you may think that it’s impossible for you to make such a mistake. But how many times did
you find yourself looking at the graded exam/problem set, realizing you made a simple mistake and you couldn’t even fathom how could you make such a mistake?
You shouldn’t underestimate the psychological factors associated with taking an exam under time constraints and knowing that the score you get will impact your grade (and potentially many other
things downstream). You can’t avoid those factors, you can only adapt to them by putting yourself in such conditions more often. Whenever you are given practice questions, put yourself in an exam
environment: set the timing, remove all distractions, and solve all problems as if it’s an actual exam. This may look like a fairly banal advice, but in many cases the key to success is in banal and
simple steps. There’s no secret formula.
As a good test for how much are you affected by psychological pressure, you can first solve all problems with time constraints, then after time runs out, take a pen of different ink and continue
solving/reviewing problems until you have the most confidence in your answers. Then grade only the things you wrote under time constraints and compare it to the score you get if you grade
everything you wrote after that. By looking at the difference you’ll see the impact of time-constraints on your performance. | {"url":"https://chem165.ischemist.com/Problem-Solving-Strategies","timestamp":"2024-11-14T03:44:23Z","content_type":"text/html","content_length":"76110","record_id":"<urn:uuid:39cf7c7d-a7c1-4d16-b304-1d58b0e2816b>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00812.warc.gz"} |
Year 9 Assessment Questions
Last week I blogged about our Year 8 assessments. I set out the principles we follow when we write our Key Stage 3 assessments, and I shared examples of the questions that challenged our highest
attaining students.
Today's post is about Year 9. My school's Year 9 cohort has a very wide range of maths attainment, which was further exacerbated by the lockdowns in Year 7 and 8. One notable feature of this year
group is how incredibly good the high attaining students are. The top set teacher finds it hard work to sufficiently challenge them in lessons. So when I made the end of year assessment, I had to
ensure there were plenty of questions in there to make them think. I don't want anyone coming out of a maths assessment bragging that they found it really easy.
Here are some of the more challenging questions from our end of Year 9 assessment.
At a concert the ratio of men to women is 5 : 3.
The ratio of women to children is 7 : 4.
Show that more than half of the people at the concert are men.
This was a middle-of-the-paper question. It's not massively challenging - I taught set four out of five and a few of them managed to get full marks on it. But I wanted to include it here because it's
a nice question requiring a bit of reasoning. It was originally from an AQA specimen GCSE paper.
A more difficult ratio question was this one from Edexcel, which also tested another Year 9 topic: changing the subject.
The ratio (y + x) : (y - x) is equivalent to k : 1.
Find a formula for y in terms of k and x.
Surface Area
Surface area is a great topic because it presents many opportunities for problem solving and reasoning. For example, this classic question may have been fairly straightforward for a high attainer,
but I like the way it's also accessible to anyone who does a bit of thinking. I was delighted that a few students in my class worked this out. We put it in our non-calculator paper so it also tested
their arithmetic.
The total surface area of a cube is 294cm^2.
Work out the volume of the cube.
A slightly more difficult question was this one from Edexcel. It doesn't have any particularly challenging reasoning in it, but has got a multiple steps to work through, including unit conversion
which might be missed. I like questions where students have to identify that they need to work with surface area rather than volume.
The most challenging surface area question I included was this one from AQA. Very few students could do this, even our highest attainers. We will revisit questions like this in Year 10.
I like this coal question from WJEC. In Year 9 we teach bounds and error intervals for the first time. We introduce some basic bounds calculations, but go into greater depth on this at GCSE. This
question fitted perfectly. However, I thought our students would find it easier than they did. I had it near the start of the paper, but most the students in my class only picked up one mark on it.
My students are all working at a Grade 4 level though. Our higher attaining students had no problem with this one.
We taught Venn Diagrams to Year 9 this year. There were some fairly straightforward Venn diagram questions nearer the start of our end of year assessment, where students basically just had to
complete various Venn Diagrams. But for challenge I wanted to test understanding of notation as well as probability, so I adapted an OCR AS level question. Only our very best mathematicians answered
this correctly.
Algebraic Proportion
Algebraic proportion can be incredibly procedural. As long as students read the question carefully, once they know how to do it, it's almost a guaranteed five marks. So I included a non-calculator
proportion question that was a bit different to questions they'd seen before. What I liked about this was the accessibility: no one from my class got all the marks, but they did manage to pick up one
or two.
Right-Angled Trigonometry
This question was my pièce de résistance. I figured that if our super clever Year 9s breezed through all the other challenging questions I threw at them, they would surely have to stop and think at
this point. This SQA question is designed to be solved using the Sine Rule. But our students don't do the Sine Rule until Year 10. This question can be done with right-angled trigonometry. The way I
did it was by splitting the base into x and 350 - x, then forming two equations for the height and equating them. Even if our students managed to get this far, solving the equation would be fairly
challenging for them because they haven't seen anything like this before.
As it happened, none of our Year 9s managed to solve it in the way I envisaged. But one very smart student came up with a genius (albeit inefficient!) method of trial and improvement. I yelped with
joy when I realised what he'd done:
It's such a delight to see students using creative approaches like this.
I also found some great challenging questions involving standard form and percentages, but I will stop at this point otherwise this blog post will go on forever! Like I said in my last post, there
are numerous places we can find good assessment questions for Key Stage 3. It's a shame they aren't centrally produced anymore - the old KS3 SATs contained great questions, but at least we can still
draw on those to make our own assessments.
If you have any good Year 9 assessment questions you'd like to share, please tweet me.
Thanks for reading! | {"url":"https://www.resourceaholic.com/2022/07/year-9-assessment-questions.html","timestamp":"2024-11-05T13:09:09Z","content_type":"application/xhtml+xml","content_length":"114614","record_id":"<urn:uuid:7bdffad5-2cee-4b58-befc-6449b86d06fa>","cc-path":"CC-MAIN-2024-46/segments/1730477027881.88/warc/CC-MAIN-20241105114407-20241105144407-00399.warc.gz"} |
Regression Archives - | Statistical Models
The post is about MCQs correlation and regression. There are 20 multiple-choice questions covering topics related to the basics of correlation and regression analysis, best-fitting trend, least
square regression line, interpretation of correlation and regression coefficients, and regression plot. Let us start with the MCQs Correlation Regression Quiz.
Online MCQs on Correlation and Regression Analysis with Answers
1. The dependent variable in a regression line is
2. The process by which we estimate the value of dependent variable on the basis of one or more independent variables is called
3. The best-fitting trend is one for which the sum of squares of error is
4. The independent variable is also called
5. The predicted rate of response of the dependent variable to changes in the independent variable is called
6. In the regression equation $Y=a+bX$, the $Y$ is called
7. If all the values fall on the same straight line and the line has a positive slope then what will be the value of the Correlation coefficient $r$:
8. For the Least Square trend $\hat{Y}=\alpha+\beta X$
9. In Regression Analysis $\sum\hat{Y}$ is equal to
10. The correlation coefficient is the _________ of two regression coefficients.
11. A relationship where the flow of the data points is best represented by a curve is called
12. In the regression equation $Y=a+bX$, the $X$ is called
13. The method of least squares directs that select a regression line where the sum of the squares of the deviations of the points from the regression line is
14. In Regression Analysis, the regression line ($Y=\alpha+\beta X$) always intersect at the point
15. The regression line always passes through
16. Which one is equal to explained variation divided by total variation?
17. In the Least Square Regression Line, $\sum(Y-\hat{Y})^2$ is always
18. If a straight line is fitted to data, then
19. All the data points falling along a straight line is called
20. In the Least Square Regression line, the quantity $\sum(Y-\hat{Y})$ is always
MCQs Correlation Regression Analysis
• In Regression Analysis $\sum\hat{Y}$ is equal to
• In the Least Square Regression Line, $\sum(Y-\hat{Y})^2$ is always
• Which one is equal to explained variation divided by total variation?
• The best-fitting trend is one for which the sum of squares of error is
• If a straight line is fitted to data, then
• In Regression Analysis, the regression line ($Y=\alpha+\beta X$) always intersect at the point
• In the Least Square Regression line, the quantity $\sum(Y-\hat{Y})$ is always
• If all the values fall on the same straight line and the line has a positive slope then what will be the value of the Correlation coefficient $r$:
• For the Least Square trend $\hat{Y}=\alpha+\beta X$
• The regression line always passes through
• The process by which we estimate the value of dependent variable on the basis of one or more independent variables is called
• The method of least squares directs that select a regression line where the sum of the squares of the deviations of the points from the regression line is
• A relationship where the flow of the data points is best represented by a curve is called
• All the data points falling along a straight line is called
• The predicted rate of response of the dependent variable to changes in the independent variable is called
• The independent variable is also called
• In the regression equation $Y=a+bX$, the $Y$ is called
• In the regression equation $Y=a+bX$, the $X$ is called
• The dependent variable in a regression line is
• The correlation coefficient is the ———– of two regression coefficients.
Best Correlation Regression Analysis MCQs 4
The post is about Correlation Regression Analysis MCQs. There are 20 multiple-choice questions. The quiz covers topics related to the coefficient of correlation, regression analysis, simple linear
regression equations, interpretation of correlation, and regression coefficients. Let us start with the Correlation Regression Analysis MCQs Quiz.
Please go to Best Correlation Regression Analysis MCQs 4 to view the test
Online Correlation Regression Analysis MCQs with Answers
• The value of the coefficient of correlation lies between
• If the scatter diagram is drawn the scatter points lie on a straight line, then it indicates
• In the model $Y= mX+ a\,\,\,$, $Y$ is also known as the:
• The regression equation is the line with a slope passing through
• If the regression equation is equal to $Y=23.6 – 54.2X$, then $23.6$ is the ———- while $-54.2$ is the ———- of the regression line.
• The sample coefficient of correlation
• If the equation of the regression line is $y = 5$, then what result will you take out from it?
• Which of the following relationships holds
• In regression equation $y=\alpha + \beta X + e$, both $X$ and $y$ variables are
• If $R^2$ is zero, that is no collinearity/ Multicollinearity, the variance inflation factor (VIF) will be
• The method of least squares finds the best fit line that ———- the error between observed & estimated points on the line
• The predicted rate of response of the dependent variable to changes in the independent variable is called
• The slope of the regression line of $Y$ on $X$ is also called
• In a simple regression, the number of unknown constants are
• In a simple regression equation, the number of variables are
• If $Y=2+0.6x$ then the value of the slope will be
• Which of the following can never be taken as the coefficient of correlation?
• When $\beta_{yx}$ is positive, then $\beta_{xy}$ will be
• If $Y=2+0.6X$ then the value of $Y$-intercept will be
• If $r=0.6$ and $\beta_{yx}=1.8$ then $\beta_{xy} = ?$
Important MCQs on Correlation and Regression 3
The post is about MCQs on Correlation and Regression Analysis with Answers. There are 20 multiple-choice questions covering the topics related to correlation and regression analysis, interpretation
of correlation and regression coefficients, relationship between variables, and correlation and regression coefficients. Let us start with MCQs on Correlation and Regression.
Please go to Important MCQs on Correlation and Regression 3 to view the test
MCQs on Correlation and Regression with Answers
• The coefficient of Correlation values lies between
• If $r_{xy} = -0.84$ then $r_{yx}=?$
• In Correlation, both variables are always
• If two variables oppose each other then the correlation will be
• A perfect negative correlation is signified by
• The Coefficient of Correlation between $U=X$ and $V=-X$ is
• The Coefficient of Correlation between $X$ and $X$ is
• The Coefficient of Correlation $r$ is independent of
• If $X$ and $Y$ are independent of each other, the Coefficient of Correlation is
• If $b_{yx} <0$ and $b_{xy} =<0$, then $r$ is
• If $r=0.6, b_{yx}=1.2$ then $b_{xy}=?$
• When the regression line passes through the origin then
• Two regression lines are parallel to each other if their slope is
• When $b_{xy}$ is positive, then $b_{yx}$ will be
• If $\hat{Y}=a$ then $r_{xy}$?
• When two regression coefficients bear the same algebraic signs, then the correlation coefficient will be
• It is possible that two regression coefficients have
• The regression coefficient is independent of
• In the regression line $Y=a+bX$
• In the regression line $Y=a+bX$ the following is always true
Best Correlation and Regression Quiz 2
The post is about the Correlation and Regression Quiz. There are 20 multiple-choice questions. The quiz covers the topics related to correlation analysis and regression analysis, Basic concepts,
assumptions, and violations of correlation and regression analysis, Model selection criteria, interpretation of correlation and regression coefficients, etc.
Please go to Best Correlation and Regression Quiz 2 to view the test
Online Correlation and Regression Quiz with Answers
• The strength (degree) of the correlation between a set of independent variables $X$ and a dependent variable $Y$ is measured by
• The percent of the total variation of the dependent variable $Y$ explained by the set of independent variables $X$ is measured by
• A coefficient of correlation is computed to be -0.95 means that
• Let the coefficient of determination computed to be 0.39 in a problem involving one independent variable and one dependent variable. This result means that
• The relationship between the correlation coefficient and the coefficient of determination is that
• Multicollinearity exists when
• If “time” is used as the independent variable in a simple linear regression analysis, then which of the following assumptions could be violated
• In multiple regression, when the global test of significance is rejected, we can conclude that
• A residual is defined as
• What test statistic is used for a global test of significance?
• If the value of any regression coefficient is zero, then two variables are said to be
• In the straight line graph of the linear equation $Y=a+bX$, the slope will be upward if
• In the straight line graph of the linear equation $Y=a+bX$, the slope will be downward if
• In the straight line graph of the linear equation $Y=a+BX$, the slope is horizontal if
• For the regression $\hat{Y}=5$, the value of regression coefficient of $Y$ on $X$ will be
• If $\beta_{yx} = -1.36$ and $\beta_{xy} = -0.34$ then $r_{xy} =$
• If one regression coefficient is greater than one then the other will be
• To determine the height of a person when his weight is given is
• The dependent variable is also called
• The dependent variable is also called | {"url":"https://itfeature.com/regression/","timestamp":"2024-11-02T22:23:15Z","content_type":"text/html","content_length":"332424","record_id":"<urn:uuid:ee600c4e-10fd-4ae8-8d63-da6882bc6f3c>","cc-path":"CC-MAIN-2024-46/segments/1730477027730.21/warc/CC-MAIN-20241102200033-20241102230033-00425.warc.gz"} |
Time Table Multiplication Worksheets
Mathematics, especially multiplication, forms the keystone of many scholastic techniques and real-world applications. Yet, for several learners, grasping multiplication can position an obstacle. To
resolve this hurdle, teachers and parents have welcomed a powerful tool: Time Table Multiplication Worksheets.
Intro to Time Table Multiplication Worksheets
Time Table Multiplication Worksheets
Time Table Multiplication Worksheets -
These Multiplication Printable Worksheets below are designed to help your child improve their ability to multiply a range of numbers by multiples of 10 and 100 mentally The following sheets develop
children s ability to use and apply their tables knowledge to answer related questions
These multiplication times table worksheets are appropriate for Kindergarten 1st Grade 2nd Grade 3rd Grade 4th Grade and 5th Grade Multiplication Times Tables Sized Chart This multiplication times
table charts is a great resource for teaching kids their multiplication times tables The chart is sized based off the magnitude of the
Significance of Multiplication Technique Comprehending multiplication is essential, laying a solid structure for sophisticated mathematical concepts. Time Table Multiplication Worksheets use
structured and targeted practice, promoting a much deeper comprehension of this fundamental arithmetic operation.
Evolution of Time Table Multiplication Worksheets
Times Tables Chart Printable 1 12 Multiplication Chart Multiplication Chart Printable
Times Tables Chart Printable 1 12 Multiplication Chart Multiplication Chart Printable
On this page you have a large selection of 2 digit by 1 digit multiplication worksheets to choose from example 32x5 Multiplication 3 Digits Times 1 Digit On these PDF files students can find the
products of 3 digit numbers and 1 digit numbers example 371x3 Multiplication 4 Digits Times 1 Digit
Here is our random worksheet generator for free multiplication worksheets The generator tests the commutative property of multiplication For example if the 3 times table is selected it will test 3 x
7 and 7 x 3 for calculations to work out Using this generator will let you create your own worksheets for Multiplying with numbers to 5x5
From traditional pen-and-paper exercises to digitized interactive layouts, Time Table Multiplication Worksheets have actually evolved, catering to diverse discovering designs and preferences.
Kinds Of Time Table Multiplication Worksheets
Standard Multiplication Sheets Simple exercises focusing on multiplication tables, aiding learners develop a solid arithmetic base.
Word Issue Worksheets
Real-life circumstances integrated into problems, enhancing vital thinking and application abilities.
Timed Multiplication Drills Tests developed to improve speed and accuracy, aiding in fast psychological math.
Benefits of Using Time Table Multiplication Worksheets
The Multiplying 1 To 12 By 9 A Math Worksheet From The Multiplicatio Multiplication Facts
The Multiplying 1 To 12 By 9 A Math Worksheet From The Multiplicatio Multiplication Facts
Method 1 To calculate an exercise involving multiplication by 7 you can calculate the same exercise but with 5 instead of 7 and add the number that appears in the exercise twice for example 3 7 3 5 3
3 15 6 21 Method 2 Multiples of 7 can be remembered in order Start from the number 7 and add 7 each time
The little diploma is made up of 30 questions Your little diploma shows you can do the 1 2 3 4 5 and 10 times tables For the big tables diploma you are given 40 questions which include all the tables
from 1 to 12 Learn the multiplication tables in an interactive way with the free math multiplication learning games for 2rd 3th 4th and 5th
Enhanced Mathematical Skills
Constant practice hones multiplication effectiveness, enhancing total math capabilities.
Boosted Problem-Solving Abilities
Word issues in worksheets develop analytical reasoning and approach application.
Self-Paced Discovering Advantages
Worksheets suit private understanding speeds, promoting a comfy and adaptable understanding environment.
Just How to Create Engaging Time Table Multiplication Worksheets
Incorporating Visuals and Colors Vivid visuals and colors capture focus, making worksheets aesthetically appealing and engaging.
Including Real-Life Scenarios
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What is the Most Useful Software in Chemical Engineering?
1. Product reviews
What is the Most Useful Software in Chemical Engineering? The List of Most Important Calculation Tools
The field of chemical engineering is in constant change, so are available calculation tools and software packages. In fast everyday life, it is a considerable challenge for a chemical engineer to
know which tool can serve best for solving a certain problem.
The different packages can be applied to solve typical problems in mass and energy balance, fluid mechanics, heat and mass transfer, unit operations, reactor engineering, and process and
equipment design and control.
In this article, we highlight the most important tools and packages with their capabilities, based on the available professional experience of an author, available literature and discussions.
The Figure below summarizes the most useful software packages in chemical engineering:
So, let's start from the beginning...
General Software for Mathematical Modeling
Python programming language
Despite starting out as a hobby project named after Monty Python, Python is now one of the most popular and widely used programming languages in the world. Besides web and software development,
Python is used for data analytics, machine learning, and even design.
Python is an object-oriented (based around data), high-level (easier for humans to understand) programming language. First launched in 1992, it’s built in a way that it’s relatively intuitive to
write and understand. As such, it’s an ideal coding language for those who want rapid development.
Python is a popular and in-demand skill to learn.
MS Excel^®
It is a known fact that Microsoft Office Excel is a spreadsheet application that features calculation, graphing tools, tables, and a macro programming language - Visual Basic. The main advantage of
Excel is that it is available and is widely used in industry and academia. Thus, it is a perfect tool or interface not only to perform calculations but also to connect different software so that the
end user can interact with Excel, and behind the scenes, other software such as CHEMCAD, MATLAB etc. is running and reporting the results back to Excel.
It is best used for:
• Built-In functions & formulas – there are a large number of built-in functions defined, such as statistics (MEAN, AVERAGE, t-test), algebraic (SUM, ROUND, LOG, LOG10), logical (IF, FALSE, etc.),
reference, database, and information. Those are easy to use in different kinds of formulas.
• Operations with columns and rows – it is easy to find & sort data and use them in replicated formulas etc.
• Plotting – there is a large number of options depending on the needs
• Solver - It is the tool to use within Excel to solve numerically a set of equations, problem optimization including fitting a set of data to a given linear and nonlinear equation and more. Solver
is an add-in that needs to be activated to be used.
• Building functions in Visual Basic for Applications - Excel has built-in capability to generate customized functions using Visual Basic for Applications (VBA). This is a powerful tool that can
save time for you without becoming an expert in programming as it opens the possibilities to run loops and conditionals on the background. This capability also allows the user to build relatively
large equations that are used in several areas of the worksheet (e.g., polynomials for the esti¬mation of specific heat of components) and allows the user to read the calculations easily when
looking at the formulas in the cells.
• Link Excel with other software - Excel has become a standard package so that a number of other specialized software use it as a source of information to report data since it is more
user-friendly. Therefore, we can use the information in Excel to be loaded in MATLAB, Hysys or CHEMCAD or transferred back to Excel.
Mathworks MATLAB^®
MATLAB is one of the most used software packages in engineering in general and also in chemical engineering. Much has been written about this popular software, more than 1500 books serving more
than 1 million users.
MATLAB is a programming language. Its operation is based on the use of .m files that can be divided in two classes, scripts and functions. A script is basically a number of operations that we want to
perform in a certain sequence. Functions are a particular type of scripts that must begin with the word “function” at the top of them. Functions can be user-defined or typical operations such as
equation solving or differential equations. Within MATLAB, we have all the algebraic, statistical functions predefined along with plotting capabilities.
MATLAB has a number of functions that allow solving linear and nonlinear equations (fzero: for one variable alone, fsolve), optimizing a function (fmincon: constrained optimization, linprog: linear
programming; fminin or fminsearch: unconstrained optimization; bintprog: binary and integer optimization), and solving differential equations (ode__) or partial differential equations (pdepe).
Some examples of how MATLAB can be used in chemical engineering include:
• Momentum, Mass, and Energy Transfer - There are a number of examples in the transport phenomena field that, even though represent different phenomena, they can be mathematically described using a
partial differential equation, the “pdepe” toolbox.
• Distillation Column Operation - McCabeMethod - typical shortcut approach for the initial conceptual estimation of the operation of binary distillation columns
• Modeling of different kinds of process equipment – heat exchangers, pumps, valves, evaporators, columns, reactors etc.
• Reactor design - The models are based on explicit algebraic equations and differential equations. Thus, we use ODEXX function in MATLAB to solve the concentration, temperature, and/or pressure
profiles along the operation of such equipment.
• Control loops analysis, control design and tuning.
Mathworks Simulink^®
Simulink® (Simulation and Link ) is a software add-on to MATLAB based on the concept of block diagrams that are common in the control engineering areas. It is an environment for dynamic simulation
and process control. Each of the blocks can contain a subsystem inside, which is helpful for big problems. We only need to select a number of blocks and with the right button of the mouse, click and
select create subsystem.
Simulink is easier to used for engineers because it does not require any programming skills, therefore models can be build using blocks instead of defining functions.
Process Simulators
The simulation, design, and optimization of a chemical process plant, which comprises several processing units interconnected by process streams, are the core activities in process engineering. These
tasks require performing material and energy balancing, equipment sizing, and costing calculation. A computer package that can accomplish these duties is known as a computer-aided process design
package or simply a process simulator.
The process simulation market underwent severe transformations in the 1985–1995 decade. Relatively few systems have survived and they inclide: CHEMCAD, Aspen Plus, Aspen HYSYS, PRO/II, ProSimPlus,
SuperPro Designer, and gPROMS.
Chemstations CHEMCAD
CHEMCAD is Chemstations’ software suite for process simulation. Features include process development, equipment design, equipment sizing, thermophysical property calculations, dynamic simulations,
process intensification studies, energy efficiency/optimization, data reconciliation, process economics, troubleshooting/process improvement, Microsoft Visual Basic etc.
The CHEMCAD suite includes six products that can be purchased individually or bundled as needed for specific industries, projects, and processes.
• CC - steady state simulations of continuous chemical processes, features libraries of chemical components, thermodynamic methods, and unit operations, enabling you to simulate processes from lab
scale to full scale. Ideal for Users who want to design processes, or rate existing processes, in steady state.
• CC – dynamics is used to conduct dynamic flowsheet analysis, operability check-out, PID loop tuning, operator training, online process control and soft sensor functionality. Ideal for users who
want to design or rate dynamic processes.
• CC-THERM is used for sizing heat exchangers, covers shell-and-tube, plate-and-frame, air-cooled, and double-pipe exchangers. Rigorous designs are based on physical property and phase equilibria
• CC-BATCH allows you to design or rate a batch distillation column.
• CC-SAFETY NET - used for analysis of any pipe network with the piping and safety relief network simulation software.
• CC – FLASH – Used to calculate physical properties and phase equilibria (VLE, LLE, VLLE) for pure components and mixtures with incredible accuracy. All products within the CHEMCAD suite feature
CC-FLASH capabilities.
ASPEN HYSYS & ASPEN PLUS
Two similar software packages with all the functionalities that process simulator should have are also the most widespread among chemical engineers. AspenTech has a wide portfolio of modeling tools,
among them most important and most known are process simulation tools Aspen Hysys and Aspen Plus.
Aspen HYSYS (or simply HYSYS) is a chemical process simulator used to mathematically model chemical processes, from unit operations to full chemical plants and refineries. HYSYS is able to perform
many of the core calculations of chemical engineering, including those concerned with mass balance, energy balance, vapor-liquid equilibrium, heat transfer, mass transfer, chemical kinetics,
fractionation, and pressure drop. HYSYS is used extensively in industry and academia for steady-state and dynamic simulation, process design, performance modeling, and optimization.
Aspen Plus is a process modeling tool for conceptual design, optimization, and performance monitoring for the chemical, polymer, specialty chemical, metals and minerals, and coal power industries. It
can also be used for mass and energy balances, physical chemistry, thermodynamics, chemical reaction engineering, unit operations, process design and process control.
In general, it can be said that Aspen Plus is better tool for chemical process design such as fine chemistry, chemicals, pharma, etc., whilst HYSYS is best for hydrocarbon, petrochemical,
petroleum operations such as natural gas, liquified gases, crude oil etc…
DWSIM - an open source process simulator
DWSIM is an open-source CAPE-OPEN compliant chemical process simulator for Windows, Linux and macOS. DWSIM is built on top of the Microsoft .NET and Mono Platforms and features a Graphical User
Interface (GUI), advanced thermodynamics calculations, reactions support and petroleum characterization / hypothetical component generation tools.
DWSIM is able to simulate steady-state, vapor–liquid, vapor–liquid-liquid, solid–liquid and aqueous electrolyte equilibrium processes with the following Thermodynamic Models and all mojor Unit
For an open sourse tool, it is a quite advanced and great support for engineers.
Specialized Software
Computational Fluid Dynamics
Computational fluid dynamics, known as CFD, is the numerical method of solving mass, momentum, energy, and species conservation equations and related phenomena on computers by using programming
CFD and multiphysics modeling and simulation can be applied to many science and engineering disciplines. The main areas in chemical engineering are the following:
• Combustion processes,
• Food process engineering,
• Fuel cells, batteries, and supercapacitors,
• Microfluidic flows and devices,
• Pipe flows and mixing,
• Reaction engineering.
The basics of CFD are partial differential equations and thus knowledge of numerical mathematics is essential to solve them with appropriate numerical technique.
Since these conservation equations are designed and solved on computers, knowledge of programming languages, such as FORTRAN, C++, Java, or MATLAB is equally important.
CFD-based software modeling tools, popular in scientific and engineering communities, are ANSYS CFX, ANSYS Fluent, ANSYS Multiphysics, COMSOL Multiphysics, FLOW-3D, STAR-CD and STAR-CCM+, and an
open-source software tool OpenFOAM. Other CFD-based software tools, such as AVL FIRE or ANSYS Polyflow, are also available on the market, but they are specialized for particular physical systems,
such as internal combustion engines, power trains, polymers, glass, metals, and cement process technologies.
The most widely used commercial software tools, such as ANSYS Fluent, STAR-CD, and STAR-CCM+, are based on finite volume method, whereas ANSYS CFX uses finite element-based control volume method. On
the other hand, COMSOL Multiphysics is based on finite element method. | {"url":"http://test.simulatelive.com/product-reviews/simulation/what-is-the-most-useful-software-in-chemical-engineering","timestamp":"2024-11-02T05:24:59Z","content_type":"text/html","content_length":"45495","record_id":"<urn:uuid:ff5c3af6-cb42-471c-b815-c775683ce3e1>","cc-path":"CC-MAIN-2024-46/segments/1730477027677.11/warc/CC-MAIN-20241102040949-20241102070949-00502.warc.gz"} |
Binary Tree Cameras
You are given the root of a binary tree. We install cameras on the tree nodes where each camera at a node can monitor its parent, itself, and its immediate children.
Return the minimum number of cameras needed to monitor all nodes of the tree.
Never expected such a simple solution to this problem. Once there is a clarity on the problem, solution may be simple. I learnt this solution from https://leetcode.com/problems/binary-tree-cameras/
discuss/1778400/C%2B%2B-Greedy-DFS-Super-Clean-Solution.-Explained-via-a-Story. He is awesome, so creative.
3 simple conditions:
— If node not exists, no camera needed
— If any one children needs a camera, place a camera on parent and increment number of cameras
— If any one children has a camera, parent does not need a camera
— By default (if above conditions not met), return camera needed
Return the values according to the above conditions
int getnumc(struct TreeNode* nd, int *nc){
/* as node is null no camera needed */
if (!nd)
return NO_CAMERA_NEEDED;
/* by default , each node needs a camera */
int lret,rret;
lret = getnumc(nd->left, nc);
rret = getnumc(nd->right, nc);
/* if any child need a camera, increment number of cameras.
place a camera
if (lret == CAMERA_NEEDED || rret == CAMERA_NEEDED){
return HAS_CAMERA;
/* if any child a camera,
parent does not need a camera */
if ((lret == HAS_CAMERA) || (rret == HAS_CAMERA))
return NO_CAMERA_NEEDED;
return CAMERA_NEEDED;
}int minCameraCover(struct TreeNode* root){
int ret, nc=0;
ret = getnumc(root, &nc);
if (ret == CAMERA_NEEDED) nc++;
One wrong approach:
I tried the below solution. Thought if left child is returning camera needed, parent has to compulsorily place a camera, So in this case why to go over the right child , just place a camera and
But missed this point : this right child may not be a single node, it can be a sub tree.
So we should be careful in optimizing the solutions :) .
#ifdef NON_WORK
/* if left child is returning camera needed,
camera will be placed at parent,
in this case, calling this func for right child is not rqd
But in this right child may be a subtree, this solution ignores going through the right subtree which is incorrect*/
int getnumc(struct TreeNode* nd, int *nc){
/* as node is null no camera needed */
if (!nd)
return NO_CAMERA_NEEDED;
/* two choices */
/* place camera on node or on children */
int lret;
lret = getnumc(nd->left, nc);
if (lret == CAMERA_NEEDED){
return HAS_CAMERA;
int rret = getnumc(nd->right, nc);
/* if any child need a camera, increment number of cameras.
place a camera
if (rret == CAMERA_NEEDED){
return HAS_CAMERA;
/* if any child a camera,
parent does not need a camera */
if ((lret == HAS_CAMERA) || (rret == HAS_CAMERA))
return NO_CAMERA_NEEDED;
return CAMERA_NEEDED; | {"url":"https://jyos-sw.medium.com/binary-tree-cameras-1a8ee3dcead1","timestamp":"2024-11-09T04:15:03Z","content_type":"text/html","content_length":"91994","record_id":"<urn:uuid:65d17f68-8728-4478-803d-a5eca55e00bb>","cc-path":"CC-MAIN-2024-46/segments/1730477028115.85/warc/CC-MAIN-20241109022607-20241109052607-00816.warc.gz"} |
Millijoule per Milligram
Units of measurement use the International System of Units, better known as SI units, which provide a standard for measuring the physical properties of matter. Measurement like latent heat finds its
use in a number of places right from education to industrial usage. Be it buying grocery or cooking, units play a vital role in our daily life; and hence their conversions. unitsconverters.com helps
in the conversion of different units of measurement like mJ/mg to MJ/kg through multiplicative conversion factors. When you are converting latent heat, you need a Millijoule per Milligram to
Megajoule per Kilogram converter that is elaborate and still easy to use. Converting mJ/mg to Megajoule per Kilogram is easy, for you only have to select the units first and the value you want to
convert. If you encounter any issues to convert Millijoule per Milligram to MJ/kg, this tool is the answer that gives you the exact conversion of units. You can also get the formula used in mJ/mg to
MJ/kg conversion along with a table representing the entire conversion. | {"url":"https://www.unitsconverters.com/en/Mj/Mg-To-Mj/Kg/Utu-8818-8802","timestamp":"2024-11-08T08:58:12Z","content_type":"application/xhtml+xml","content_length":"111769","record_id":"<urn:uuid:ea59a821-7dc4-467a-af18-8c7bf13d9560>","cc-path":"CC-MAIN-2024-46/segments/1730477028032.87/warc/CC-MAIN-20241108070606-20241108100606-00896.warc.gz"} |
Anthony Morse Explained
Death Place: Alameda County, California
Birth Date: 21 August 1911
Birth Place: Ithaca, New York
Anthony Perry Morse (21 August 1911 - 6 March 1984) was an American mathematician who worked in both analysis, especially measure theory, and in the foundations of mathematics. He is best known as
the co-creator, together with John L. Kelley, of Morse - Kelley set theory. This theory first appeared in print in Kelley's General Topology.^[1] Morse's own version appeared later in A Theory of
Sets.^[2] ^[3]
He is also known for his work on the Morse - Sard theorem and the Federer–Morse theorem.
Anthony Morse should not be confused with Marston Morse, known for developing Morse theory.
He received his PhD in 1937 at Brown University with C. R. Adams as thesis advisor. After two years at the Institute for Advanced Study he joined the mathematics faculty at Berkeley where except for
two interruptions he worked for the rest of his life on mathematics. In the first of these, from 1943 until the end of World War II, he worked on ballistics at the Aberdeen Proving Ground. In 1950
his life was interrupted by the McCarthyist loyalty oath controversy. He was one of 29"non-signers".^[4] But he was also one of 6 who took advantage of a 10-day grace period to sign, while continuing
to refer tothe remaining non-signers as "patriots."^[5]
His doctoral students include Herbert Federer, Woody Bledsoe, and Maurice Sion.
External links
Notes and References | {"url":"https://everything.explained.today/Anthony_Morse/","timestamp":"2024-11-03T15:07:34Z","content_type":"text/html","content_length":"9182","record_id":"<urn:uuid:bf471771-0ba3-4b10-a0a1-cad2a28aba5a>","cc-path":"CC-MAIN-2024-46/segments/1730477027779.22/warc/CC-MAIN-20241103145859-20241103175859-00132.warc.gz"} |
Single-Qubit Quantum Gates on a Bloch Sphere
Using the Bloch sphere, a cubit
can be represented as a unit vector (shown in red) from the origin to the point on the unit sphere with spherical coordinates . A single-qubit quantum gate operating on produces a rotated qubit , rep
resented by the green vector. Check the box for "add gate 2?" to perform a second operation using gate . This produces another qubit , which is represented by the blue vector. You can choose from the
gates H, X, Y, Z, S and T as defined in the Details. | {"url":"https://www.wolframcloud.com/objects/demonstrations/SingleQubitQuantumGatesOnABlochSphere-source.nb","timestamp":"2024-11-12T03:50:30Z","content_type":"text/html","content_length":"277813","record_id":"<urn:uuid:adc6ce86-19b1-4273-a4d7-dad420313e93>","cc-path":"CC-MAIN-2024-46/segments/1730477028242.50/warc/CC-MAIN-20241112014152-20241112044152-00001.warc.gz"} |
Re: st: Use extended functions outside of macro assignment?
Notice: On April 23, 2014, Statalist moved from an email list to a forum, based at statalist.org.
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Re: st: Use extended functions outside of macro assignment?
From James Sams <[email protected]>
To [email protected]
Subject Re: st: Use extended functions outside of macro assignment?
Date Wed, 7 Sep 2011 02:45:14 -0500
On Wednesday, September 07, 2011, you wrote:
> The logic of -list- with -if- is like any other command. Consider
> list if 2 == 2
> 2 == 2 is (vacuously) true when considered for observation 1, for
> observation 2, and so on, so every variable and every observation will
> be -list-ed. The same will be true of your syntax. If your condition
> is true, i.e. there is a match, then everything will be listed.
No, you are misunderstanding. Take the latter example, id is a variable, ids is
a local macro list. The conditional would be re-evaluated for each
observation's value of the 'id' variable. Same for the regexm example.
(repasted for reference).
> > Something like this:
> > list if regexm("`: label (varname)'", "my_regex")
> > list if `: list posof id in ids'
Doing this in a two step process often gets tedious due to the size of the
data-set and memory limitations. Being able to operate on value labels directly
as though they were the value of the variable would be particularly helpful.
Also, list was an example. I am rarely interested in actually listing data.
I've got a class of problems that I am trying to solve and being able to
dynamically invoke extended functions in the manner described would be helpful.
If it is not possible, so be it. But I wanted to make sure my question was
clear first. I thought I had seen someone doing something very close to this at
some point but cannot find it again.
* For searches and help try:
* http://www.stata.com/help.cgi?search
* http://www.stata.com/support/statalist/faq
* http://www.ats.ucla.edu/stat/stata/ | {"url":"https://www.stata.com/statalist/archive/2011-09/msg00217.html","timestamp":"2024-11-14T03:33:50Z","content_type":"text/html","content_length":"11336","record_id":"<urn:uuid:0f9f623a-cd80-4f6c-991e-44dd9e8b28cb>","cc-path":"CC-MAIN-2024-46/segments/1730477028526.56/warc/CC-MAIN-20241114031054-20241114061054-00152.warc.gz"} |